Four small spheres each of radius 'r' and mass 'm' are placed with their centers at the four corners of a square of side 'b'. The M.I. of the system about any side of square is 

Two spheres sitting on an axis have a moment or inertia of$2\times \frac{2}{5}{\mathrm{mR}}^2=\frac{4}{5}{\mathrm{mR}}^2$

By using parallel axis theorem,

Moment of Inertia of two spheres at a distance of b resting on the edge parallel to the axis is

 $2\left(\frac{2}{5}{\mathrm{mR}}^2+{\mathrm{mb}}^2\right)=\frac{4}{5}{\mathrm{mR}}^2+2{\mathrm{mb}}^2$
Total moment of inertia is hence

$\frac{4}{5}{\mathrm{mR}}^2+2{\mathrm{mb}}^2+\frac{4}{5}{\mathrm{mR}}^2=\frac{8}{5}{\mathrm{mR}}^2+2{\mathrm{mb}}^2$

Hence the correct answer is $\frac{8{\mathrm{mr}}^2}{5}+2{\mathrm{mb}}^2$.