\( {}^{238}_{92}A \rightarrow {}^{234}_{90}B + {}^{4}_{2}D + Q \)
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of \( {}^{238}_{92}A = 238.05079 \times 931.5 \, \text{MeV}/c^{2} \),
Mass of \( {}^{234}_{90}B = 234.04363 \times 931.5 \, \text{MeV}/c^{2} \),
Mass of \( {}^{4}_{2}D = 4.00260 \times 931.5 \, \text{MeV}/c^{2} \)]