A thick-walled hollow sphere has outside radius $$R_0$$. It rolls down an incline without slipping and its speed at the bottom is $$v_0$$. Now the incline is waxed, so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be $$5v_0/4$$. The radius of gyration of the hollow sphere about an axis through its centre is

Let the mass of the sphere be $$M$$ and the vertical height of the incline be $$h$$ (same in both experiments).

Case 1: Rolling without slipping (rough incline)
For pure rolling, translational speed $$= v_0$$ and angular speed $$\omega = \dfrac{v_0}{R_0}$$.
Total kinetic energy at the bottom: $$K_{\text{total}} = \tfrac12 M v_0^{2} + \tfrac12 I \omega^{2} = \tfrac12 M v_0^{2} + \tfrac12 \bigl(Mk^{2}\bigr)\Bigl(\dfrac{v_0}{R_0}\Bigr)^{2} = \tfrac12 M v_0^{2}\Bigl(1 + \dfrac{k^{2}}{R_0^{2}}\Bigr).$$

This comes from the loss of potential energy: $$Mgh = \tfrac12 M v_0^{2}\Bigl(1 + \dfrac{k^{2}}{R_0^{2}}\Bigr). \qquad -(1)$$

Case 2: Sliding without rotation (smooth, waxed incline)
With negligible friction, the sphere does not roll, so it acquires only translational kinetic energy. Given speed at the bottom is $$v_s = \dfrac{5v_0}{4}$$.
Thus $$Mgh = \tfrac12 M v_s^{2} = \tfrac12 M\Bigl(\dfrac{5v_0}{4}\Bigr)^{2} = \tfrac12 M v_0^{2}\,\dfrac{25}{16}. \qquad -(2)$$

The left sides of (1) and (2) are identical (same mass and height), so equate the right sides:

$$\tfrac12 M v_0^{2}\Bigl(1 + \dfrac{k^{2}}{R_0^{2}}\Bigr) = \tfrac12 M v_0^{2}\,\dfrac{25}{16}.$$

Cancel the common factor $$\tfrac12 M v_0^{2}$$ to get $$1 + \dfrac{k^{2}}{R_0^{2}} = \dfrac{25}{16}.$$

Solve for $$k^2$$: $$\dfrac{k^{2}}{R_0^{2}} = \dfrac{25}{16} - 1 = \dfrac{9}{16} \;\;\Longrightarrow\;\; k = \dfrac{3R_0}{4}.$$

Hence the radius of gyration of the thick-walled hollow sphere is $$\boxed{\dfrac{3R_0}{4}}$$.

Option B which is: $$3R_0/4$$