Spheres snugly fitted into a cube

If it is a 2D problem,
A circle is tangent to the two adjacent sides of a square. Its center will be on the diagonal from the corner. It needs a length of
$(\sqrt2+1)*R$ from the corner. However it spares a legth of $(\sqrt2-1)*R$ from the corner, for other circles. This is used by a smaller circle radius r, as $(\sqrt2+1)*r$ .
That is $(\sqrt2-1)*R=(\sqrt2+1)*r.\ \ \implies\ \ \dfrac R r=\dfrac{\sqrt2+1}{\sqrt2-1}.$
If there are n smaller and smaller circles, the ratio of original to the n-th circle will be .... $\dfrac R r=(\dfrac{\sqrt2+1}{\sqrt2-1})^n.$
If it is 3D with sheres in a cube. $\sqrt2 \ is\ replaced\ by \sqrt3.$ , since the distance of centers of the circles from the corner is now
from $\sqrt2R\ \ to\ \ \sqrt3R.\\ So\ answer\ to\ our\ problem \ is \left \{ \dfrac{\sqrt3+1}{\sqrt3 - 1} \right \}^2=13.92820.$

Let $R, C, S$ = Radius, distance of center and distance of sphere from the cube vertex

For any sphere $C = \sqrt{3} R$

$C_B + R_B = S_A$ which gives $R_B = R_A \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Similarly, $R_C = R_B \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = R_A \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Hence, $\frac{R_A}{R_C} = (\frac{\sqrt{3} + 1}{\sqrt{3} - 1})^2 = \textbf 13.9282$

@Harish Chandra Rajpoot @Niranjan Khanderia

In general, the radius $r_n$ of nth sphere is given as

$r_n=R(2-\sqrt{3})^{n-1}$

hence, for third sphere $r_3=R(2-\sqrt{3})^2$

$\frac{R}{r_3}=\frac{1}{(2-\sqrt{3})^2}\approx 13.9282$