I came across this problem and it seems easy but I may be missing something;
Here is my attempt. If you consider the cross-section of this 3D shape through the centres of the spheres,
The ratio of r:R = 1:2, hence linear scale factor = 2.
Therefore, volume scale factor = 8
I am then stuck here... I would appreciate any thoughts on how to proceed from here or any other alternative approach.
Let $s_U, s_D$ be the upper and lower region of the smaller sphere cut out by the plane, respectively, and $S_U, S_D$ be the upper and lower region of the larger sphere cut out by the plane, respectively.
There are several ways to prove that $s_U$ is similar to $S_D$, and $s_D$ is similar to $S_U$ (for instance, by looking at the cross sections on the plane perpendicular to the plane in the graph and passing through both centers, everything spans down to planar geometry).
Since the radii have the ration $1:2$, this tells us
$$ \begin{aligned} Vol(s_U) &= \frac18 Vol(S_D);\\ Vol(s_D) &= \frac18 Vod(S_U). \end{aligned} $$
Together with the given fact that
$$ Vol(s_U) + Vol(S_U) = \frac12\left(Vol(s_D) + Vol(S_D)\right), $$ everything reduces to a system of linear equations.