‌12H+‌12H→‌23‌He+‌01n+3.27‌MeV
The given fusion reaction is:

‌12H+‌12H→‌23‌He+‌01n+3.27‌MeV
Amount of deuterium, m=2‌kg
1 mole, i.e., 2g of deuterium contains 6.023×1023 atoms.

So, 2.0‌kg of deuterium contains ‌

6.023×1023

2

×2000=6.023×1026‌ atoms ‌
Two atoms of deuterium fuse to release 3.27‌MeV energy.

So, total energy released
‌=‌

3.27

2

×6.023×1026‌MeV
‌=‌

3.27

2

×6.023×1026×106×1.6 ×10−19J
‌=15.75×1013J
Power of the electric lamp, P=800W=800J∕ s Hence, the energy consumed by the lamp per second =800J
So, the electric lamp will glow for
‌

15.75×1013

800

s‌=0.0197×1013 s
=‌

0.0197×1012

60×60×24×365

‌=6246.8‌ years ‌