Let $m=5\times 10^{-3}$ kg, $R=4/3\times 10^{-2}$ m, $r=2/3\times 10^{-2}$ m, and $J=\sqrt{\pi /2}\times 10^{-2}$ N$\cdot$s.

Translation: $J=mv\Rightarrow v=J/m$. Time to return to the start under gravity is $t=2v/g=2J/(mg)$.

Rotation: The impulse acts at a perpendicular distance $r$ from the diameter through the centre, so the angular impulse is $Jr$ and

$Jr=I_d\omega$, with $I_d=\frac{1}{4}m{R}^2$ (moment of inertia of a thin disc about a diameter).

$\therefore \omega =\frac{4Jr}{m{R}^2}$.

Number of rotations:

$n=\frac{\omega t}{2\pi }=\frac{1}{2\pi }\cdot \frac{4Jr}{m{R}^2}\cdot \frac{2J}{mg}=\frac{4J^{2}r}{\pi m^2{R}^{2}g}$.

Plug in the numbers:

$J^2=\frac{\pi }{2}\times 10^{-4}$,$r=\frac{2}{3}\times 10^{-2}$,$m^2=25\times 10^{-6}$,$R^2=\frac{16}{9}\times 10^{-4}$,$g=10$.

$n=\frac{4\cdot \frac{\pi }{2}\times 10^{-4}\cdot \frac{2}{3}\times 10^{-2}}{\pi \cdot 25\times 10^{-6}\cdot \frac{16}{9}\times 10^{-4}\cdot 10}=30$.

$\therefore n=30$.