Ans.

Four spheres, each of mass m and radius a, are placed at the corners of the square of side b, as shown in Fig.1.

We know that the moment of inertia of the sphere about its diameter is \( \frac{2}{5}ma^2 \), [ To know the derivation of the moment of inertia of a sphere about its diameter (CLICL HERE) ].

We want to calculate the moment of inertia of the system about the axis EH.

The moment of inertia of each of the sphere E and H about the axis EH is
\( \frac{2}{5}ma^2 \).

The moment of inertia of each of the sphere F and G about he axis EH is
\( \left(\frac{2}{5}ma^2+mb^2\right) \).

So the moment of inertia of the whole system about the side EH is

\( 2\times{\frac{2}{5}ma^2}+2\times{\left(\frac{2}{5}ma^2+mb^2\right)} \)
= \( \frac{8}{5}m{a^2}+2mb^2 \)
= \( \frac{2}{5}m\left(4a^2+5b^2\right) \)