A solid sphere is rolling on a horizontal plane without slipping....

The angular momentum is given by

$L = {I}_{CM}\omega = \dfrac{2M{R}^{2}\omega}{5}$

The total kinetic energy is

$E = \dfrac{1}{2} ( M{v}^{2} + \dfrac{2M{R}^{2}{\omega}^{2}}{5} )$

The ratio of angular momentum and kinetic energy is

$\dfrac{L}{E} = \dfrac{{I}_{CM}\omega}{\dfrac{1}{2}M{v}^{2} + \dfrac{1}{2} \cdot \dfrac{2}{5}M{R}^{2}{\omega}^{2}} = \dfrac{\dfrac{2}{5}M{R}^{2}\omega}{\dfrac{7}{10}M{R}^{2}{\omega}^{2}}$

$⇒ \dfrac{\pi}{22} = \dfrac{4}{7} \cdot \dfrac{1}{\omega}$$⇒ \omega = 4rad{s}^{- 1}$