A star initially has  ${10}^{40}$ deuterons. It produces energy via the processes ${}_{1}H^2+{}_{1}H^2\to {}_{1}H^3+p$ and  ${}_{1}H^2+{}_{1}H^3\to {}_{2}H{e}^4+n$ . If the average power radiated by the star is ${10}^{16} \mathrm{W}$ , the deuteron supply of the star is exhausted in a time of the order of:
The mass of the nuclei are as follows:

$\mathrm{M}\left({\mathrm{H}}^2\right)=2.014 \mathrm{amu}$;  $\mathrm{M}\left(\mathrm{n}\right)=1.008 \mathrm{amu}$ ; $\mathrm{M}\left(p\right)=1.008 \mathrm{amu}$; $\mathrm{M}\left({\mathrm{He}}^4\right)=4001 \mathrm{amu}$
 

The given reactions are:
${}_{1}H^2+{}_{1}H^2\to {}_{1}H^3+p$
${}_{1}H^2+{}_{1}H^3\to {}_{2}H^4+n$
$3{}_{1}H^2\to {}_{2}H{e}^4+n+p$
Mass defect
$\Delta m=(3\times 2.014-4.001-1.007-1.008)amu$
$=0.026amu$
Energy released 
$=0.026\times 931MeV$
$=0.026\times 931\times 1.6\times {10}^{-13}J$

$=3.87\times {10}^{-12}J$
This is the energy produced by the consumption of three deuteron atoms.
Total energy released by deuterons
$=\frac{{10}^{40}}{3}\times 3.87\times {10}^{-12}J$
$=1.29\times {10}^{28}J$
The average power radiated is $\mathrm{P}={10}^{16} \mathrm{W}$ or ${10}^{16} J/\mathrm{s}$  .
Therefore, total time to exhaust all deuterons of the star will be
$t=\frac{1.29\times {10}^{28}}{{10}^{16}}=1.29\times {10}^{12} s\approx {10}^{12} s$