(d) For a disc moment of inertia about a tangential axis in its own plane $= \frac{5}{4}MR^{2}$
$\therefore$ $M_{1}K_{1}^{2} = \frac{5}{4}M_{1}R^{2}$
$\Rightarrow$ $K_{1} = \frac{\sqrt{5}}{2}\ R$
Now, for a ring moment of inertia about a tangential axis in its own plane $= \frac{3}{2}\ M_{2}R^{2}$
$\therefore$ $M_{2}K_{2}^{2} = \frac{3}{2}\ M_{2}R^{2}$
$\Rightarrow$ $K_{2} = \sqrt{\frac{3}{2}}\ R$
$\therefore$ $\frac{K_{1}}{K_{2}} = \frac{\sqrt{5}}{\sqrt{6}}$