Assuming that about 20 MeV of energy is released per fusion reaction

${}_1{\mathrm{H}}^2 + {}_1{\mathrm{H}}^2 \to { }_0{\mathrm{n}}^1 +{ }_2{\mathrm{He}}^4$

Then the mass of ${}_1\mathrm{H}^2$  consumed per day in a fusion reactor of power 1 megawatt will approximately be

$$\begin{gathered} \mathrm{P}={10}^6\mathrm{watt} \\ \mathrm{Time} 1 \mathrm{day}=24 \mathrm{x}36 \mathrm{x}{10}^2 \sec \end{gathered}$$

Energy produced, 

$$\begin{gathered} \mathrm{U}={\mathrm{P}}_{\mathrm{t}}={10}^6 \mathrm{x} 24 \mathrm{x} 36 \mathrm{x} {10}^2 \\ =24 \mathrm{x} 36 \mathrm{x} {10}^8 \mathrm{joule} \end{gathered}$$

Energy released for fusion reaction

=20 Mev

$=32 \mathrm{x} {10}^{-13}\mathrm{joule}$

Energy released per fusion

$=32 \mathrm{x} {10}^{-13}\mathrm{joule}$

Number of  fusion

$$\begin{gathered} =\frac{24 \mathrm{x} 36 \mathrm{x} {10}^8}{32 \mathrm{x} {10}^{-13}}=27 \mathrm{x} {10}^{21} no of {}_1{}^{ }H^2 atoms involved \\ 2\times 27\times {10}^{21} \\ \mathrm{Mass} \mathrm{of} 6 \mathrm{x} {10}^{23}\mathrm{atoms} = 2\mathrm{g} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Mass} \mathrm{of} 2\times 27 \mathrm{x} {10}^{21} \mathrm{atoms} \\ =\frac{2}{6 \mathrm{x} {10}^{23}}2\mathrm{x} 27 \mathrm{x} {10}^{21}=0.2 \mathrm{g} \end{gathered}$$