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