Explanation

(c) The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$I_{CM} = \frac{1}{2}MR^{2}$
where $M$ is the mass of disc and $R$ its radius. According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc about an axis touching the disc at its diameter and normal to the disc is
$I = I_{CM} + MR^{2}$
$= \frac{1}{2}MR^{2} + MR^{2}$
$= \frac{3}{2}MR^{2}$