Five Squares in a Circle

Let the radius of the circle be $1$ . Then the side length of the big square is $\sqrt{1^2+1^2} = \sqrt 2$ . Let the side length of the small square be $a$ . Then we note that

$\begin{aligned} \left(\frac 1{\sqrt 2} + a\right)^2 + \left(\frac a2 \right)^2 & = 1^2 \\ \frac 12 + \sqrt 2 a + a^2 + \frac {a^2}4 & = 1 \\ \frac 54a^2 + \sqrt 2 a & = \frac 12 \\ a^2 + \frac {4\sqrt 2}5 a & = \frac 25 \\ \left(a + \frac {2\sqrt 2}5\right)^2 & = \frac 25 + \frac 8{25} = \frac {18}{25} \\ a + \frac {2\sqrt 2}5 & = \frac {3\sqrt 2}5 \\ \implies a & = \frac {\sqrt 2}5 \end{aligned}$

Then the ratio of areas $\dfrac {(\sqrt 2)^2 + 4a^2}{\pi \cdot 1^2} = \dfrac {2+4 \cdot \frac 2{25}}\pi = \dfrac {58}{25\pi}$ . Therefore the required answer is $58+25^3 = \boxed{15683}$ .

Let side of large and small squares are $l$ and $x$ respectively, then: $\\ l = \sqrt {2}r \\ r^2 = (\dfrac {l}{2} + x)^2 + (\dfrac {x}{2})^2 \\ r^2 = (\dfrac {\sqrt {2}}{2}r + x)^2 + (\dfrac {x}{2})^2 \\ 5x^2 + 4\sqrt 2xr - 2r^2 = 0 \\ \implies x = \dfrac {\sqrt 2}{5} r \\ S_{squares} = l^2 + 4x^2 = \dfrac {58}{25}r^2 \\ \dfrac {S_{squares}}{S_{circle}} = \dfrac {\dfrac {58}{25}r^2}{\pi r^2} = \dfrac {58}{25\pi} \\ \implies a= 58 \quad b = 25 \\ \therefore a + b^3 = \boxed{15683}$