When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E which is Translational is

Let's consider that the radius is R and the mass is M.
MOI =$\mathrm{I}=\frac{2}{3}{\mathrm{MR}}^2$  for the moment of inertia.

Rolling without slipping is a hollow sphere. Where v is the speed of the center of mass and is the angular speed, $v=R\omega$.

Translational kinetic energy = $\frac{1}{2}{\mathrm{Mv}}^2$

Rotational Kinetic energy $=\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}\left(\frac{2}{3}{\mathrm{MR}}^2\right){\left(\frac{\mathrm{v}}{\mathrm{R}}\right)}^2=\frac{1}{3}{\mathrm{Mv}}^2$

Total kinetic energy = $\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{3}{\mathrm{Mv}}^2=\frac{5}{6}{\mathrm{Mv}}^2$

Percentage of translational energy ,

$\frac{\frac{1}{2}{\mathrm{Mv}}^2}{\frac{5}{6}{\mathrm{Mv}}^2}\times 100=60\%$

Hence the correct answer is 60%.