A solid sphere of mass $M$ and radius $R$ rolls without slipping on a horizontal surface with translational speed $v$. The total kinetic energy of the sphere is:

Total KE = translational + rotational: $\dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2$. For a solid sphere $I = \dfrac{2}{5}MR^2$ and the rolling constraint gives $\omega = v/R$, so the rotational piece is $\dfrac{1}{2} \cdot \dfrac{2}{5}MR^2 \cdot \dfrac{v^2}{R^2} = \dfrac{1}{5}Mv^2$. Sum: $\dfrac{1}{2}Mv^2 + \dfrac{1}{5}Mv^2 = \dfrac{7}{10}Mv^2$. Option A ignores the rotational contribution.