${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\rightarrow {}_2^3\mathrm{He}+{}_0^1\mathrm{n}+3.27 \text{ MeV}$ The given fusion reaction is: ${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\rightarrow {}_2^3\mathrm{He}+{}_0^1\mathrm{n}+3.27 \text{ MeV}$ Amount of deuterium, $m=2 \text{ kg}$ 1 mole, i.e., $2 \text{ g}$ of deuterium contains $6.023 \times 10^{23}$ atoms. So, $2.0 \text{ kg}$ of deuterium contains $\; \frac{6.023 \times 10^{23}}{2}$ $\times 2000 = 6.023 \times 10^{26} \; \text{ atoms } \;$ Two atoms of deuterium fuse to release $3.27 \text{ MeV}$ energy. So, total energy released $\; = \; \frac{3.27}{2} \times 6.023 \times 10^{26} \text{ MeV}$ $\; = \; \frac{3.27}{2} \times 6.023 \times 10^{26} \times 10^6 \times 1.6$ $\times 10^{-19} \text{ J}$ $\; = 15.75 \times 10^{13} \text{ J}$ Power of the electric lamp, $P = 800 \text{ W} = 800 \text{ J} / \text{ s}$ Hence, the energy consumed by the lamp per second $= 800 \text{ J}$So, the electric lamp will glow for$\; \frac{15.75 \times 10^{13}}{800} \text{ s} \; = 0.0197 \times 10^{13} \text{ s}$$= \; \frac{0.0197 \times 10^{12}}{60 \times 60 \times 24 \times 365} \; = 6246.8 \; \text{ years } \;$