Number of atoms in $2 \mathrm{g}$ of ${}_1{}^{2}A=6.023\times {10}^{23}$

$\therefore$ No of atoms in $2 \mathrm{kg}$ of ${}_1{}^{2}A=\frac{6.023\times {10}^{23}}{2}\times 2000$

$\because$ Energy released in the fusion of two ${}_1{}^{2}A$ nuclei $=3.27\mathrm{MeV}$

$\therefore$ Energy released in the fusion of $2 \mathrm{kg}$ of ${}_1{}^{2}A$

$\therefore E=\frac{327}{2}\times 6.023\times {10}^{26}\times 16\times {10}^{-13}\mathrm{J}$

$=15.76\times {10}^{13}\mathrm{J}$

Power of bulb, $P=100\mathrm{W},=100$ $\mathrm{J} {\mathrm{s}}^{-1}$

$\therefore$ Time for which the bulb will glow $t=\frac{E}{P}$

$=\frac{15.75\times {10}^{13}}{100}$

$=15.75\times {10}^{11}\mathrm{s}$

$=\frac{15.75\times {10}^{11}}{60\times 60\times 24\times 365}$

$=5\times {10}^4$ years