Two identical solid spheres each of mass 2   kg and radii 10   cm   are fixed at the ends of…

The moment of inertia of a sphere is given by the formula $I=\frac{2}{5}M{R}^2$.

Since there are two masses, the total moment of inertia about the axis perpendicular to the rod passing through the centre is 

$I=2\times \left[\frac{2}{5}M{R}^2+M{\left(\frac{l}{2}+r\right)}^2\right] ...\left(i\right)$

The given data is 

$$\begin{gathered} M=2 \mathrm{kg} \\ l=\left(0.4-0.2\right) \mathrm{m}=0.2 \mathrm{m} \\ \mathrm{R}=0.1 \mathrm{m} \end{gathered}$$

Substituting the values in equation (i)

$$\begin{gathered} I=\left(\frac{4}{5}(2\times 0.1^2)+2\times 2(\frac{0.2}{2}+0.1{)}^2\right) \mathrm{kg} {\mathrm{m}}^2 \\ \Rightarrow I=\left(\left(\frac{4}{5}\times 0.02\right)+\left(4\times 0.04\right)\right) \mathrm{kg} {\mathrm{m}}^2 \\ \Rightarrow I=\left(0.016+0.16\right) \mathrm{kg} {\mathrm{m}}^2=0.176 \mathrm{kg} {\mathrm{m}}^2 \end{gathered}$$