Four spheres each of diameter ' 2   a ' and mass ' m ' are placed in a way that their…

Let $AB$ is the axis about which we have to find the moment of inertia, the moment of inertia of the system is the summation of the moment of inertia of each sphere about $AB$axis, $I=I_A+I_B+I_C+I_D$

$I_A=I_B=\frac{2}{5}m{a}^2$

Using the parallel axis theorem for the sphere $C \text{and} D$

$$\begin{gathered} \therefore I \text{of }C \text{about} AB=I \text{of }C\text{ about} CD+m{b}^2 \\ {I}_C=I_D=\frac{2}{5}m{a}^2+m{b}^2 \end{gathered}$$

So, $I=I_A+I_B+I_C+I_D$

$$\begin{gathered} I=\frac{2}{5}m{a}^2+\frac{2}{5}m{a}^2+\left(\frac{2}{5}m{a}^2+m{b}^2\right)+\left(\frac{2}{5}m{a}^2+m{b}^2\right) \\ I=\frac{8}{5}m{a}^2+2m{b}^2 \end{gathered}$$