The moment of inertia of a sphere about its central axis and a solid spherical shell of mass M and radius R as shown in the figure above is $I_{\text{sphere}} = \large\frac{2}{5}$$MR^2$

Kinetic energy of the rotation of the sphere KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 $ where $\omega $ is the angular velocity.

$\Rightarrow$ KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 = \large\frac{1}{2} \frac{2}{5}$$ MR^2 \omega^2$

Now, angular velocity $\omega = \large\frac{v}{R}$, where $v$ is the linear velocity of the sphere.

$\Rightarrow$ KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 = \large\frac{1}{2} \frac{2}{5}$$ MR^2 $$\large\frac{v^2}{R^2}$$ = \large\frac{1}{5} $$Mv^2$

Kinetic energy of the linear motion KE$_{\text{linear motion}} = \large\frac{1}{2}$$Mv^2$

Total Kinetic energy $=$ KE$_{\text{rotation}} + $KE$_{\text{linear motion}} = \large\frac{1}{5}$$Mv^2 + \large\frac{1}{2}$$ Mv^2$$ = \large\frac{7}{10} $$Mv^2$

Now, the fraction of its total energy associated with rotation $= \large\frac {\text{KE of Rotation}}{\text{Total KE}}$$ = \Large\frac{ \large\frac{1}{5} \normalsize Mv^2}{\large\frac{7}{10} \normalsize Mv^2}$

$\qquad = \large\frac{2}{7}$