A hollow sphere of radius R rests on a smooth horizontal surface....

Let the mass of the hollow sphere be $m$ and the applied horizontal force be $F$.

The linear acceleration of the center of mass is given by Newton's second law:

$a = \dfrac{F}{m}$

The torque acting about the center of mass due to the force at the highest point is:

$\tau = F R$

The moment of inertia of a hollow sphere about its center of mass is $I = \dfrac{2}{3}mR^2$. The angular acceleration $\alpha$ is:

$\alpha = \dfrac{\tau}{I} = \dfrac{FR}{\dfrac{2}{3}mR^2} = \dfrac{3F}{2mR}$

Using the kinematic equations for linear and angular motion starting from rest, the distance travelled $s$ and the angular displacement $\theta$ in time $t$ are:

$s = \dfrac{1}{2}at^2$

$\theta = \dfrac{1}{2}\alpha t^2$

Taking the ratio of $s$ to $\theta$:

$\dfrac{s}{\theta} = \dfrac{a}{\alpha} = \dfrac{\dfrac{F}{m}}{\dfrac{3F}{2mR}} = \dfrac{2R}{3}$

For one full rotation, the angular displacement is $\theta = 2\pi$. Substituting this into the ratio gives the distance travelled:

$s = \dfrac{2R}{3} \times 2\pi = \dfrac{4\pi R}{3}$