Explanation

(a) The rotational kinetic energy of the disc is
$K_{rot} = \frac{1}{2}I\omega^{2} = \frac{1}{2}\left( \frac{1}{2}MR^{2} \right)\omega^{2} = \frac{1}{4}MR^{2}\omega^{2}$
The translational kinetic energy is
$K_{trans} = \frac{1}{2}Mv_{CM}^{2}$
where $v_{CM}$ is the linear velocity of its centre of mass.
Now, $v_{CM} = R\omega$
Therefore, $K_{trans} = \frac{1}{2}MR^{2}\omega^{2}$
Thus, $K_{total} = \frac{1}{4}MR^{2}\omega^{2} + \frac{1}{2}MR^{2}\omega^{2} = \frac{3}{4}\ MR^{2}\omega^{2}$
$\therefore$ $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{4}MR^{2}\omega^{2}}{\frac{3}{4}\ MR^{2}\omega^{2}} = \frac{1}{3}$