The given reactions are:

${}_1{\mathrm{H}}^2 +{ }_1{\mathrm{H}}^2 \to { }_1{\mathrm{H}}^3 + \mathrm{p}$

${}_1{\mathrm{H}}^2 +{ }_1{\mathrm{H}}^3 \to { }_1{\mathrm{He}}^4 + \mathrm{n}$

$3_1{\mathrm{H}}^2\to { }_2{\mathrm{He}}^4 +\mathrm{n} + \mathrm{p}$

Mass defect

$\mathrm{\Delta m} = (3 \times 2. 014 - 4. 001 - 1. 007 - 1. 008) \mathrm{amu}$
$= 0.026 \mathrm{amu}$

Energy released$= 0. 026 \times 931 \mathrm{MeV}$

$= 0. 026 \times 931 \times 1. 6 \times {10}^{-12}\mathrm{J} = 3. 87 \times {10}^{-13} \mathrm{J}$

This is the energy produced by the consumption of three deuteron atoms.

Therefore Total energy released by 10⁴⁰ deuterons

$=\frac{10^{40}}{3}$$\times 3. 87 \times {10}^{-12} \mathrm{J} = 1. 29 \times 1028 \mathrm{J}$

The average power radiated is $P={10}^{16} \mathrm{W}$ or ${10}^{16} \mathrm{J} \mathrm{s}$. Therefore, the total time to exhaust all deuterons of the star, time taken will be

$t=\genfrac{}{}{0.1ex}{}{1.29\times 10^{28}}{10^{16}}=1.29\times \mathrm{}\mathrm{}$$10^{12} \approx 10^{12}$