One quarter of the disc of mass $m$  is removed. If $r$  be the radius of the disc, the new moment of inertia is

Momentum of inertia whole disc about an axis through centre of disc and perpendicular to its plane is $\text{I}=\frac{1}{2}m{r}^2$

As one quarter of disc is removed, new mass,

$m\text{'}=\frac{3}{4}m\therefore \frac{1}{2}\left(\frac{3}{4}m\right)r^2=\frac{3}{8}m{r}^2$ .