a
Mass per unit area of disc \(= \frac{M}{{\pi {R^2}}}\)

Mass of removed portion of disc, 

\(M' = \frac{M}{{\pi {R^2}}} \times \pi {\left( {\frac{R}{2}} \right)^2} = \frac{M}{4}\)

Moment of inertia of removed portion about an axis passing through center of disc \(O\) and perpendicular to the plane od disc,

\(I{'_0} = {I_0} + M'{d^2}\)

\(= \frac{1}{2} \times \frac{M}{4} \times {\left( {\frac{R}{2}} \right)^2} + \frac{M}{4} \times {\left( {\frac{R}{2}} \right)^2}\)
\(= \frac{{M{R^2}}}{{32}} + \frac{{M{R^2}}}{{16}} = \frac{{3M{R^2}}}{{32}}\)

When portion of disc would not have been removed, the moment of inertia of complete disc about center \(O\) is 

\({I_0} = \frac{1}{2}M{R^2}\)

So, moment of inertia of the disc with removed portion is 

\(I = {I_0} - {I_0} = \frac{1}{2}M{R^2} - \frac{{3M{R^2}}}{{32}} = \frac{{13M{R^2}}}{{32}}\)