Let the mass of the thin uniform disc be $M$ and its radius be $R$. The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is $I_1 = \frac{1}{2}MR^2$. If $K_1$ is the corresponding radius of gyration, then $MK_1^2 = \frac{1}{2}MR^2 \implies K_1 = \frac{R}{\sqrt{2}}$. The moment of inertia of the disc about its diameter is $I_2 = \frac{1}{4}MR^2$. If $K_2$ is the corresponding radius of gyration, then $MK_2^2 = \frac{1}{4}MR^2 \implies K_2 = \frac{R}{2}$. The ratio of their radii of gyration is: $\frac{K_1}{K_2} = \frac{\frac{R}{\sqrt{2}}}{\frac{R}{2}} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{1} = \sqrt{2}:1$.