A solid sphere of mass ' m ' and radius ' R ' is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with angular speed twice that of sphere. The ratio of kinetic energy of sphere to kinetic energy of cylinder will be

Given a solid sphere and solid cylinder with equal mass $m$ and radius $R$, rotating about their respective axes with angular speeds $\omega_c = 2\omega_s$.

The moment of inertia for a solid sphere is $I_s = \frac{2}{5}mR^2$, yielding rotational kinetic energy $KE_s = \frac{1}{2}I_s\omega_s^2 = \frac{1}{5}mR^2\omega_s^2$.

For the solid cylinder, $I_c = \frac{1}{2}mR^2$, and with $\omega_c = 2\omega_s$, its kinetic energy becomes $KE_c = \frac{1}{2}I_c\omega_c^2 = mR^2\omega_s^2$.

The ratio is therefore $\frac{KE_s}{KE_c} = \frac{\frac{1}{5}mR^2\omega_s^2}{mR^2\omega_s^2} = \frac{1}{5}$.

Answer: $\boxed{B}$