A disc of mass m and radius R has a concentric hole of radius r. Its M.I. about an axis through centre and normal to its plane is

Moment of inertia of the disc with hole:

${\mathrm{I}}_0=\frac{1}{2}{\mathrm{m}}_1{\mathrm{R}}^2-\frac{1}{2}{\mathrm{m}}_2{\mathrm{r}}^2$

${\mathrm{m}}_1-{\mathrm{m}}_2=\mathrm{m} .....\left(1\right)$

$\mathrm{also}, {\mathrm{m}}_1\propto {\mathrm{R}}^2, {\mathrm{m}}_2\propto {\mathrm{r}}^2$

$\therefore \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2} ....\left(2\right)$

using (1) and (2),  

${\mathrm{m}}_1\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)=\mathrm{m}$

$\Rightarrow {\mathrm{m}}_1=\frac{{\mathrm{mR}}^2}{{\mathrm{R}}^2-{\mathrm{r}}^2} , {\mathrm{m}}_2=\frac{{\mathrm{mr}}^2}{{\mathrm{R}}^2-{\mathrm{r}}^2}$

$\therefore {\mathrm{I}}_0=\frac{\mathrm{m}}{2\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}\left({\mathrm{R}}^4-{\mathrm{r}}^4\right)$$=\frac{m}{2}\left(R^2+r^2\right)$