Radius of gyration is given by, $K=\sqrt{\frac{I}{m}}$
We know moment of inertia for the ring $I_{\text {disc }}=\frac{1}{2} m R^2$ and the disc $I_{\text {ring }}=m R^2$ about its central axis perpendicular to its plane.
Using parallel axis theorem, the moment of inertia about a tangential axis perpendicular to the plane can be easily obtained using,
$I^{\prime}=I+m R^2$
The radius of gyration is defined as:
$\begin{aligned}
& K=\sqrt{\frac{I}{m}} \\
& \therefore K_{\text {disc }}=\sqrt{\frac{\frac{1}{2} m R^2+m R^2}{m}}=\sqrt{\frac{3}{2}} R \\
& \therefore K_{\text {ring }}=\sqrt{\frac{m R^2+m R^2}{m}}=\sqrt{2} R
\end{aligned}$
Thus, the ratio of the radius of gyration of a ring to that of a disc of same radius and mass, about a tangential axis perpendicular to the plane is:
$\frac{K_{\text {disc }}}{K_{\text {ring }}}=\frac{\sqrt{3 / 2}}{\sqrt{2}}=\frac{\sqrt{3}}{2}$