A star has \(100 \%\) helium composition. It starts to convert three \({ }^4 \mathrm{He}\) into one \({ }^{12} \mathrm{C}\) via triple alpha process as \({ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}\). The mass of the star is \(2.0 \times 10^{32} \mathrm{~kg}\) and it generates energy at the rate of \(5.808 \times 10^{30} \mathrm{~W}\). The rate of converting these \({ }^4 \mathrm{He}\) to \({ }^{12} \mathrm{C}\) is \(\mathrm{n} \times 10^{42} \mathrm{~s}^{-1}\), where \(\mathrm{n}\) is _______. [Take, mass of \({ }^4 \mathrm{He}=4.0026 \mathrm{u}\), mass of \({ }^{12} \mathrm{C}=12 \mathrm{u}\)]

1. A \(14\)
2. B \(5\)
3. C \(15\)
4. D \(20\)