Ans.

Let us consider a spherical ball of radius r and mass M.
If v be the linear velocity of the ball then the translational kinetic energy of the ball is \( \frac{1}{2}Mv^2 \).
If \( \omega \) be the angular velocity of the ball then the rotational kinetic energy of the ball is \( \frac{1}{2}I{\omega}^2 \), [ To know the derivation of the rotational kinetic energy of a body, (CLICK HERE) ].
Where \( I=\frac{2}{5}Mr^2 \) is the moment of inertia of the spherical ball about its diameter. [ To know the derivation (CLICK HERE) ].

So the total energy of the spherical ball is

\( E \)=\( \frac{1}{2}Mv^2+\frac{1}{2}I{\omega}^2\\=\frac{1}{2}Mv^2+\frac{1}{2}\frac{2}{5}Mr^2{\omega}^2\\=\frac{1}{2}Mv^2+\frac{1}{5}Mv^2\\=\frac{7}{10}Mv^2 \)

[where, \( v=r\cdot\omega \)]

\( \frac{Translational\ K.E.}{Rotational\ K.E.} \) \( =\displaystyle{\frac{\frac{1}{2}Mv^2}{\frac{1}{5}Mv^2}}\\=\frac{5}{2} \)

\( \frac{Rotational\ K.E.}{Total\ K.E.} \) \( = \displaystyle{\frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2}}\\=\frac{2}{7} \)