Mass per unit area of the original disc is $\sigma = \frac{M}{\pi R^2}$. The diameter of the cut-out hole is $R$, so its radius is $\frac{R}{2}$. Mass of the cut-out portion, $m' = \sigma \times \text{Area} = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$. The hole's rim passes through the centre of the original disc, which means the distance between the centre of the original disc and the centre of the cut-out portion is $d = \frac{R}{2}$. Moment of inertia of the original disc about an axis passing through its centre and perpendicular to its plane is $I_0 = \frac{1}{2}MR^2$. Moment of inertia of the cut-out portion about an axis passing through its own centre and perpendicular to its plane is $I'_{cm} = \frac{1}{2}m'\left(\frac{R}{2}\right)^2 = \frac{1}{2}\left(\frac{M}{4}\right)\left(\frac{R^2}{4}\right) = \frac{MR^2}{32}$. Using the parallel axis theorem, the moment of inertia of the cut-out portion about the centre of the original disc is $I'_0 = I'_{cm} + m'd^2 = \frac{MR^2}{32} + \left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$. Moment of inertia of the remaining part is $I_{rem} = I_0 - I'_0 = \frac{1}{2}MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.