To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere (K₁) to the radius of gyration for a thin hollow sphere (K₂) of the same mass M and radius R.

The moment of inertia (I) of a solid sphere about its own axis is given by :

$$I_{solid} = \frac{2}{5}MR^2$$

The radius of gyration (K) is related to the moment of inertia (I) and mass (M) by the formula :

$I = MK^2$

So for the solid sphere, we can find K₁ using :

$$K^2_{1} = \frac{I_{solid}}{M} = \frac{2}{5}R^2$$

$$K_{1} = R\sqrt{\frac{2}{5}}$$

Now, for a thin hollow sphere, the moment of inertia about its axis is given by :

$$I_{hollow} = \frac{2}{3}MR^2$$

We can find K₂ using :

$$K^2_{2} = \frac{I_{hollow}}{M} = \frac{2}{3}R^2$$

$$K_{2} = R\sqrt{\frac{2}{3}}$$

Now, we need to find the ratio K₁ : K₂ :

$$\frac{K_{1}}{K_{2}} = \frac{R\sqrt{\frac{2}{5}}}{R\sqrt{\frac{2}{3}}}$$

The R terms cancel out, and we are left with :

$$\frac{K_{1}}{K_{2}} = \frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}}$$

Simplify by taking the square root of the fraction :

$$\frac{K_{1}}{K_{2}} = \frac{\sqrt{2}\sqrt{3}}{\sqrt{5}\sqrt{2}}$$

Now, the square root of 2 terms cancel out, giving us :

$$\frac{K_{1}}{K_{2}} = \frac{\sqrt{3}}{\sqrt{5}}$$