1. Write the nuclear reaction

   Proton capture by lithium gives: ${}_3^7\text{Li} + {}_1^1\text{H} \rightarrow 2\,{}_2^4\text{He} + Q$

2. Calculate mass defect per reaction

   Initial mass: $m_i = m({}_3^7\text{Li}) + m({}_1^1\text{H}) = 7.0160 + 1.0079 = 8.0239\,u$

   Final mass: $m_f = 2\times m({}_2^4\text{He}) = 2\times 4.0026 = 8.0052\,u$

   Mass defect: $\Delta m = m_i - m_f = 8.0239 - 8.0052 = 0.0187\,u$

3. Convert mass defect into energy per reaction

   Given: $1\text{ nucleon mass} \approx 1u \approx 1\,\text{GeV}/c^2$

   Hence, $Q = 0.0187\,\text{GeV} = 18.7\,\text{MeV}$

   In joules: $18.7\,\text{MeV} = 18.7\times 10^6 \times 1.6\times 10^{-19}$ $Q \approx 2.992\times 10^{-12}\,\text{J}$

4. Find number of lithium nuclei in 20 g of ${}^7\text{Li}$

   Molar mass of ${}^7\text{Li}$ is approximately $7\,\text{g/mol}$.

   Number of moles in $20\,\text{g}$: $n = \frac{20}{7}\,\text{mol}$

   Number of nuclei: $N = \frac{20}{7}N_A = \frac{20}{7}\times 6.02\times 10^{23}$ $N \approx 1.72\times 10^{24}$

5. Total energy released

   $E = NQ = 1.72\times 10^{24}\times 2.992\times 10^{-12}$ $E \approx 5.15\times 10^{12}\,\text{J}$

6. Convert joules to kWh

   $1\,\text{kWh} = 3.6\times 10^6\,\text{J}$

   Therefore, $E = \frac{5.15\times 10^{12}}{3.6\times 10^6} \,\text{kWh}$ $E \approx 1.43\times 10^6\,\text{kWh}$

7. Match with the nearest option

   The computed value is closest to: $1.33\times 10^6\,\text{kWh}$

   So the correct option is D.

Note: Using more accurate constants gives about $1.39\times 10^6\,\text{kWh}$, but among the given choices, D is the best match.