${}_{92}{}^{238} \mathrm{A}\to {}_{90}{}^{234} \mathrm{B}+{}_2{}^4\mathrm{D}+\mathrm{Q}$

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of ${}_{92}{}^{238} \mathrm{A}=238.05079\times 931.5 \mathrm{MeV} {\mathrm{c}}^{-2}$, mass of  ${}_{90}{}^{234}B=234.04363\times 931.5 \mathrm{MeV} {\mathrm{c}}^{-2}$, mass of ${}_2{}^4\mathrm{D}=4.00260\times 931.5\mathrm{Me}V {\mathrm{c}}^{-2}$ ]