The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is:

For a disc about the axis through its centre, normal to its plane: $I_1 = \frac{mR^2}{2}$ About its diameter: $I_2 = \frac{mR^2}{4}$ Radius of gyration $k = \sqrt{\frac{I}{m}}$ $\frac{k_1}{k_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{mR^2/2}{mR^2/4}} = \sqrt{2}$ So $k_1 : k_2 = \sqrt{2} : 1$