For a first order reaction, the ratio of time for the completion of 99.9% and half of the reaction is

For reaction first order, ${\mathrm{t}}_{1/2}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\frac{\mathrm{a}}{2}}$

$\begin{array}{l}{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2.303}{\mathrm{k}}\log 2=\frac{0.693}{\mathrm{k}}\\ {\mathrm{t}}_{99.9\mathrm{\%}}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-0.999\mathrm{a}}\\ =\frac{2.303}{\mathrm{k}}\log {10}^3=\frac{2.303}{\mathrm{k}}\times 3\\ \frac{{\mathrm{t}}_{99.9\mathrm{\%}}}{{\mathrm{t}}_{1/2}}=\frac{6.9}{0.693}\approx 10.\end{array}$