Four spheres of diameter 2a and mass M are placed with their centers on the four corners of a square of side 'b' . Then the moment of inertia of the system about an axis along one of the sides of the square is 

MOI (moment of inertia) of sphere about mass centre: $\frac{2}{5}{\mathrm{Ma}}^2$

By parallel axis theorem,

$\mathrm{I}={\mathrm{I}}_{\mathrm{cm}}+{\mathrm{Mx}}^2$ 

Where x is the distance between the centre of mass and the axis of rotation and $I_{cm}$ is the MOI around the centre of mass.

System MOI relative to a square's side,

$=2\left(\frac{2}{5}{\mathrm{Ma}}^2\right)+2(\frac{2}{5}{\mathrm{Ma}}^2+{\mathrm{Mb}}^2)=\frac{8}{5}{\mathrm{Ma}}^2+2{\mathrm{Mb}}^2$

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