Energy of photon is given by
$$
E=\frac{h c}{\lambda}
$$
where $h$ is the Planck's constant, $c$ the velocity of light and $\lambda$ its wavelength. de-Broglie wavelength is given by
$$
\lambda=\frac{h}{p}
$$
where $p$ is being momentum of photon. From Eqs. (i) and (ii), we get
$$
E=\frac{h c}{h / p}=p c
$$
Or $p=E / c$
$$
\text { Given, } \begin{aligned}
&=1 \mathrm{MeV}=1 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}, \\
c &=3 \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, after putting numerical values, we obtain
$$
\begin{aligned}
p &=\frac{1 \times 10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^{8}} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\
&=5 \times 10^{-22} \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
$$