Given, $M=200 \mathrm{MeV} {\mathrm{c}}^{-2}$ and $m=1 \mathrm{GeV} {\mathrm{c}}^{-2}$, now use the concept of conservation of linear momentum, $M{v}_0=mv\Rightarrow v=\frac{M{v}_0}{m} ...\left(1\right)$.

Now use the conservation of energy, 

$\frac{1}{2}M{\left(v_0\right)}^2=\frac{1}{2}m{\left(v\right)}^2+∆E ..\left(2\right)$, here $∆E=13.6 \mathrm{eV}$, for the hydrogen atom in the ground state, now from equation, $\left(1\right) \& \left(2\right)$, $\frac{1}{2}M{\left(v_0\right)}^2=\frac{1}{4}\left(3∆E\times \frac{10}{8}\right) e\mathrm{V}=\frac{N}{4}$,

$\Rightarrow \frac{N}{4}=\frac{51}{4} e\mathrm{V}$$\Rightarrow N=51$.