Answer
2.224 MeV.
Work Step by Step
The neutron capture described is $^{1}_{1}H+\;^{1}_{0}n\rightarrow\;^{2}_{1}H $.
Assume that the initial KE is nearly zero. The energy of the gamma ray is the energy release from the difference in the masses, using data from Appendix B.
$$E_{\gamma}=\left( m(^{1}_{1}H)+ m(^{1}_{0}n)-m(^{2}_{1}H) \right)c^2$$
$$ =\left(1.007825u+1.008665u-2.014102u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{\gamma}=2.224\;MeV$$
