Let the angular speed of the solid sphere be $\omega$. Then the angular speed of the solid cylinder is $2\omega$.

The moment of inertia of a solid sphere about its diameter is $I_{\text{sphere}} = \frac{2}{5}mR^2$. Its rotational kinetic energy is $E_{\text{sphere}} = \frac{1}{2}I_{\text{sphere}}\omega^2 = \frac{1}{2} \left(\frac{2}{5}mR^2\right) \omega^2 = \frac{1}{5}mR^2\omega^2$.

The moment of inertia of a solid cylinder about its geometrical axis is $I_{\text{cylinder}} = \frac{1}{2}mR^2$. Its rotational kinetic energy is $E_{\text{cylinder}} = \frac{1}{2}I_{\text{cylinder}}(2\omega)^2 = \frac{1}{2} \left(\frac{1}{2}mR^2\right) (4\omega^2) = mR^2\omega^2$.

The ratio of their kinetic energies of rotation is: $\frac{E_{\text{sphere}}}{E_{\text{cylinder}}} = \frac{\frac{1}{5}mR^2\omega^2}{mR^2\omega^2} = \frac{1}{5} = 1:5$.