A solid spherical ball is rolling on a frictionless horizontal...

Under pure rolling condition: $v = \omegaR$

Translational $K . E = \dfrac{1}{2}m{v}^{2}$

Rotational $K . E = \dfrac{1}{2} \times \dfrac{2}{5}m{R}^{2}{\left(\dfrac{v}{R}\right)}^{2} = \dfrac{1}{5}m{v}^{2}$

Total $K . E = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{5}m{v}^{2} = \dfrac{7}{10}m{v}^{2}$

So, $\dfrac{RotationalK \cdot E}{TotalK \cdot E} = \dfrac{\dfrac{1}{5}m{v}^{2}}{\dfrac{7}{10}m{v}^{2}} = \dfrac{2}{7}$