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

Given: $$^{238}_{92}A \to ^{234}_{90}B + ^{4}_{2}D + Q$$

Mass of A = $$238.05079 \times 931.5$$ MeV/c$$^2$$

Mass of B = $$234.04363 \times 931.5$$ MeV/c$$^2$$

Mass of D = $$4.00260 \times 931.5$$ MeV/c$$^2$$

$$\Delta m = m_A - m_B - m_D = (238.05079 - 234.04363 - 4.00260) \times 931.5$$ MeV/c$$^2$$

$$= (238.05079 - 238.04623) \times 931.5 = 0.00456 \times 931.5$$ MeV/c$$^2$$

$$Q = \Delta m \cdot c^2 = 0.00456 \times 931.5 = 4.248$$ MeV $$\approx 4.25$$ MeV

The correct answer is Option D: $$4.25$$ MeV.