${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\to {}_2^3\mathrm{He}+{}_0^1\mathrm{n}+3.27\mathrm{MeV}$
The given fusion reaction is:

${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\to {}_2^3\mathrm{He}+{}_0^1\mathit{n}+3.27\mathrm{MeV}$
Amount of deuterium, $\mathit{m}=2\mathrm{kg}$
1 mole, i.e., $2\mathrm{g}$ of deuterium contains $6.023\times {10}^{23}$ atoms.

So, $2.0\mathrm{kg}$ of deuterium contains $\frac{6.023\times {10}^{23}}{2}$

Two atoms of deuterium fuse to release $3.27\mathrm{MeV}$ energy.

So, total energy released
$=\frac{3.27}{2}\times 6.023\times {10}^{26}\mathrm{MeV}$
$=\frac{3.27}{2}\times 6.023\times {10}^{26}\times {10}^6\times 1.6$ $\times {10}^{-19}\mathrm{J}$
$=15.75\times {10}^{13}\mathrm{J}$
Power of the electric lamp, $\mathit{P}=800\mathrm{W}=800\mathrm{J}∕\mathrm{}\mathit{s}$ Hence, the energy consumed by the lamp per second $=800\mathrm{J}$
So, the electric lamp will glow for
$\frac{15.75\times {10}^{13}}{800}\mathrm{}\mathit{s}=0.0197\times {10}^{13}\mathrm{}\mathit{s}$