The energy in MeV is released due to transformation of 1 kg mass completely into energy $(\mathrm{c} = 3 \mathrm{x} {10}^8 \mathrm{m}/\mathrm{s})$

$$\begin{gathered} \mathrm{E}=∆{\mathrm{mc}}^2=1 \mathrm{x} {(3 \mathrm{x} {10}^8)}^2=9 \mathrm{x} {10}^{16 }\mathrm{J} \\ \Rightarrow \mathrm{E}=\frac{9 \mathrm{x} {10}^{16}}{1.6 \mathrm{x} {10}^{-19}}=5.625 \mathrm{x} {10}^{35} \mathrm{eV} \\ =5.625 \mathrm{x} {10}^{29} \mathrm{M} \mathrm{eV}. \end{gathered}$$