Calculate rate constant of a first order reaction having pre exponential factor $1.6 \times 10^{13} \mathrm{~s}^{-1} \quad\left(\mathrm{E}_{\mathrm{a}} / 2.303 \mathrm{RT}=21\right)$

The rate constant $k$ for a first-order reaction follows the Arrhenius equation: $k = A e^{-E_a / RT}$.

Taking natural logarithms yields $\ln k = \ln A - \frac{E_a}{RT}$, which converts to base 10 as $\log k = \log A - \frac{E_a}{2.303 RT}$.

Given $A = 1.6 \times 10^{13} \text{ s}^{-1}$ and $\frac{E_a}{2.303 RT} = 21$, substitution gives $\log k = \log (1.6 \times 10^{13}) - 21$.

Evaluating the logarithm: $\log (1.6 \times 10^{13}) = \log 1.6 + \log 10^{13} \approx 0.2041 + 13 = 13.2041$.

Thus, $\log k = 13.2041 - 21 = -7.7959$.

Taking the antilogarithm: $k = 10^{-7.7959} \approx 1.6 \times 10^{-8} \text{ s}^{-1}$, which corresponds to option D.

Final answer: $\boxed{\text{D}}$