For rolling spherical shell, the ratio of rotational kinetic energy...

The rotational kinetic energy is given by

${K}_{rot} = \dfrac{I{\omega}^{2}}{2} = \dfrac{1}{3}M{R}^{2}{\omega}^{2}$.

The total kinetic energy is given by

${K}_{tot} = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{3}M{R}^{2}{\omega}^{2} = \dfrac{m{R}^{2}{\omega}^{2}}{2} + \dfrac{1}{3}M{R}^{2}{\omega}^{2} = \dfrac{5}{6}M{R}^{2}{\omega}^{2}$

Thus, the ratio is

$\dfrac{{K}_{rot}}{{K}_{tot}} = \dfrac{\dfrac{1}{3}M{R}^{2}{\omega}^{2}}{\dfrac{5}{6}M{R}^{2}{\omega}^{2}} = \dfrac{2}{5}$

Comparing with the given value,

$\dfrac{x}{5} = \dfrac{2}{5} ⇒ x = 2$