When a solid sphere rolls without slipping on a rough horizontal surface, the ratio of its translational kinetic energy to its total kinetic energy is

Ratio of Translational to total kinetic energy of a rolling body is

$K{E}_{Translation}:K{E}_{total}=1:\left(1+\frac{k^2}{R^2}\right)$

We know, $k=\sqrt{\frac{2}{5}}R$ for a solid sphere. 

$\therefore K{E}_{Translation}:K{E}_{total}=1:\left(1+\frac{2}{5}\right)=5:7$