A particular reaction has a rate constant of $1.3\times {10}^{-2}{ \mathrm{s}}^{-1}.$ Find out the ratio of $\frac{{\mathrm{t}}_{1/2}}{{\mathrm{t}}_{99.9}}$

$$\begin{gathered} \mathrm{K}=1.3\times {10}^2{ \mathrm{s}}^{-1} \\ {\mathrm{t}}_{1/2} \mathrm{so} 1\mathrm{st} \mathrm{order} \\ \begin{array}{rl} & {\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ & {\mathrm{t}}_{99.2\mathrm{\%}}=\frac{2.303}{\mathrm{k}}\log \left(\frac{100}{0.8}\right)\end{array} \\ \frac{{\mathrm{t}}_{1/2}}{\mathrm{t}99.2\mathrm{\%}}=\frac{\frac{6.693}{\mathrm{k}}}{\frac{2.303}{\mathrm{k}}{\displaystyle \log }{\displaystyle \left(\frac{100}{0.8}\right)}} \\ \frac{{\displaystyle 0.693}}{{\displaystyle 2.303\times \log \left(125\right)}} \\ =\frac{{\displaystyle 0.693}}{{\displaystyle 2.303\times 2.09}}\Rightarrow 1/10 \end{gathered}$$

Therefore, the correct option is (B).