The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I_{CM} = \frac{1}{2}MR^2$. According to the theorem of parallel axes, the moment of inertia about an axis touching the edge of the disc and normal to its plane is given by: $I = I_{CM} + Md^2$ Here, the distance between the two parallel axes is $d = R$. $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$