Answer
$2/7$
Work Step by Step
The sphere is rotating with the axis at its center, so $I=\frac{2}{5}MR^2$
The rotational kinetic energy of the sphere is
$$KE_R=\frac{1}{2}\times\frac{2}{5}MR^2\omega^2=\frac{1}{5}MR^2\omega^2$$
The sphere's linear speed is also its tangential speed at the outer edge, so we have $$v=R\omega$$
The total KE is $$KE=\frac{1}{2}Mv^2+KE_R=\frac{1}{2}MR^2\omega^2+\frac{1}{5}MR^2\omega^2$$ $$KE=\frac{7}{10}MR^2\omega^2$$
Thus: $$\frac{KE_R}{KE}=\frac{2}{7}$$
