The given fusion reaction is:

$_{1}^{2} H+_{1}^{2} H \rightarrow_{2}^{3} H e+n+3.27 M e V$

Amount of deuterium, $m=2 kg$

1 mole, i.e., 2 g of deuterium contains $6.023 \times 10^{23}$ atoms.

$\therefore 2.0 kg$ of deuterium contains $=\frac{6.023 \times 10^{23}}{2} \times 2000=6.023 \times 10^{26}$ atoms

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

$\therefore$ Total energy per nucleus released in the fusion reaction:

$E=\frac{3.27}{2} \times 6.023 \times 10^{26} MeV$

$=\frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^{6}$

$=1.576 \times 10^{14} J$

Power of the electric lamp, $P=100 W =100 J / s$

Hence, the energy consumed by the lamp per second = $100 J$ The total time for which the electric lamp will glow is calculated as

$\frac{=1.576 \times 10^{14}}{100} s$

$\frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^{4}\; years$