It's beautiful

Consider the top view of the setup. The projection of a circle will be an ellipse. Since the circles are tilted at pi/4 the semi major axis will remain 10 while the semi minor axis will be 10 tan pi/4. now we have three ellipses tangent to each other and a circle passing through them as shown.

notice that the slope of the drawn tangent is tan pi/6. Using some basic trigonometry and calculus we can find the side of the triangle and hence the radius of the circle, which comes out to be $\sqrt{160/3}$

It is easy to realize that three circles are in-circles of three triangles. These triangles are three faces of the pyramid as in the diagram. $HK=a*\frac{\sqrt{3}}{6}$ $\angle{SKH}=45^{0}$ $SA=\sqrt{\frac{a^2}{4}+\frac{a^2}{6}}=a\sqrt{\frac{5}{12}}$ $r(a+2a\sqrt{\frac{5}{12}})=a*\frac{a}{\sqrt{6}}$ $a=10*(\sqrt{6}*(1+2*\frac{5}{12}))$ $SA=10*(\sqrt{6}*(1+2*\frac{5}{12}))*\frac{5}{12}$ $\frac{PQ}{a}=\frac{SA-\frac{a}{2}}{SA}$ $A=R_{PQR}=\frac{PQ}{\sqrt{3}}=7.3$ $[10A]=\boxed{73}$

The given plane is the $Oxz$ plane, and the circles are placed symmetrically around the origin. One of the circles has its center at $x = 0$ . The coordinates of this center are $(0, r/\sqrt2, c)$ , where $c$ is yet to be determined.

The points on this circle may be parametrized as $(x,y,z) = (r\cos\theta, \frac r{\sqrt 2}(1-\sin\theta), c - \frac r{\sqrt 2}\sin\theta).$ The direction of the tangent to the circle at each point is $(dx,dy,dz) \propto (-\sin\theta, \frac{\cos\theta}{\sqrt 2}, \frac{-\cos\theta}{\sqrt 2}).$

Because of symmetry, we expect this circle to touch neighboring circles in the planes $x/z=\pm\tan 60^\circ = \pm\sqrt 3$ . The tangent vectors at the point of contact must also lie in this plane, $dx/dz = \pm\sqrt 3$ . We focus on the positive side. This gives the equations $\sqrt 3= \frac x z = \frac{r\cos\theta}{c - (r\sqrt 2)\sin\theta} \ \ \therefore\ \ c-\frac{r\sin\theta}{\sqrt 2} = \frac{r\cos \theta}{\sqrt 3},$ $\sqrt 3 = \frac{dx}{dz} = \frac{-\sin\theta}{-(\cos\theta)/\sqrt 2}\ \ \therefore\ \ \frac{\sin\theta}{\cos\theta} = \frac{\sqrt 3}{\sqrt 2}.$ Thus we find $\theta = \text{inv tan}\frac{\sqrt 3}{\sqrt 2} \approx 50.77^\circ;\ \ \ \ \sin\theta=\frac{\sqrt 3}{\sqrt 5}, \cos\theta = \frac{\sqrt2}{\sqrt 5}.$ The coordinates of the points where the circles touch are $(x,y,z) = (r\cos\theta, \frac r{\sqrt 2}(1-\sin\theta), \frac{r\cos \theta}{\sqrt 3}) = r\ (\frac{\sqrt 2}{\sqrt 5}, \frac{1}{\sqrt2}-\frac{\sqrt 3}{\sqrt{10}}, \frac{\sqrt 2}{\sqrt{15}}).$ Finally, the radius of the circle through the touching points is $A = \sqrt{x^2 + z^2} = r\ \sqrt{\frac 2 5 + \frac 2{15}} = r\ \sqrt{\frac 8{15}} \approx 0.7303\ r.$ Substituting $r = 10$ we get $\lfloor 10A \rfloor=73$ .

The three circumferences are inscribed into three triangles, in particular let's consider three triangles such that they define a tetrahedron whose faces are inclined of 45° respect the base plane. Let's call AOB one of this three triangles who contains the three circles. Be h the tetrahedron's height.
Being the circumferences incliend of 45° the height of the triangle AOB results to be h*sqrt(2). I've preferred handwrite the solution .

The projection of the three circles on the plane would be three ellipses.The vertical semi axis will be contracted to $\text{The projection of the three circles on the plane would be three ellipses.}\\ \text{The vertical semi axis will be contracted to } 10Cos45=\dfrac{10}{\sqrt2}.\\ \color{#D61F06}{Ellipse_1}\\ Ellipse_1\ \text{ is along X-axis with center at (0,0). Its equation is } \color{#3D99F6}{ 2X^2+(\pm Y)^2=100.}\\ Slope ~ of~one\ of\ its\ tangent~ ST_3,\ is\ m_3=\dfrac{dy}{dx}.:-\\ Where~4X+( - Y)\dfrac{dy}{dx}=0, \color{#3D99F6}{m_3=\dfrac{dy}{dx}= \dfrac{2X} Y.}\\ \text{Since 3 circles are placed symmetric w.r.t. vertical line of symmetry through S, }\\ \text{tangent lines } ST_2,ST_3,\ of\ Ellipse_1~ \text{would be at an angle of 360/3=120.}\\ \therefore\ Slope\ of\ ST_3,~is~m_3= Tan(\dfrac{120}{2} ) = \sqrt3 .\ But\ \color{#3D99F6}{ m_3 = \dfrac {2X} Y }\ also. \\ \therefore~X=- \dfrac{\sqrt3*Y} 2. ~\text{Putting this X value of point of tangency} ~T_3 ~ in\ equation\ for\ Ellipse_1,\\ 2* \left ( \dfrac{\sqrt3*Y} 2 \right )^2 +Y^2=100, ~and~solving~we~get~Y=2\sqrt{10}.\\ \therefore\ T_2T_3 =2*Y=4\sqrt{10}. ~~~~~\\ \color{#D61F06}{All\ three\ Ellipses}\\ \text{For the 3 ellipses, these three lines joining 2 points of tangency of an ellipse,}\\ \text{form an equilateral triangle } T_1T_2T_3. \\ \text{The circumcircle of this triangle passes through the three points of tangency. }\\ \therefore~ST_1=R=\dfrac{4\sqrt{10}}{\sqrt3}=7.30=A\\ \therefore~\lfloor {10A} \rfloor= \Large~~\color{#D61F06}{73}$

Let the plane that the circles are inclined to be the xy-plane. Circles of radius $10$ are then projected to ellipses of semi-major axis length $10$ and semi-minor axis length $5\sqrt{2}$ . Set the origin to be the center of symmetry of this setup. Suppose one of these ellipses in centered at $(a, 0)$ . By symmetry, this ellipse should be tangent to the lines $y=\pm\tan{(60^\circ )}x=\pm\sqrt{3}x$ . The ellipse can be described by the parametric equations $x=a+5\sqrt{2}\cos{t}, y=10\sin{t}.$ It follows that the gradient of the tangent to a point $(x,y)$ is $\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{100\cos{t}}{-50\sin{t}} = \frac{2(x-a)}{-y}.$ It is expected that if $y=\sqrt{3}x$ is substituted to this expression, the gradient value should be $\sqrt{3}$ : $\sqrt{3} = \frac{dy}{dx} = \frac{2(x-a)}{-\sqrt{3}x} \Rightarrow 3x=2(a-x) \Rightarrow x=\frac{2a}{5} \Rightarrow y=\frac{2a\sqrt{3}}{5}.$ Substituting these values into the ellipse equation $2(x-a)^2+y^2=100$ and solving for $a$ gives us $a=5\sqrt{\frac{10}{3}}$ (the negative solution gives a symmetric case. So the radius of the circle is simply $A=\frac{y}{\cos{(30^\circ )}}$ , leading to $\lfloor10A\rfloor =\lfloor\frac{40a}{5}\rfloor =\boxed{73}$ .