Time required for completion of 99 . 9 % of first order reaction is ________ times of half life ( t 1 / 2 )…

For first order reaction:
$\mathrm{kt}=\log \frac{\left[{\mathrm{A}}_0\right]}{\left[{\mathrm{A}}_{\mathrm{t}}\right]}$

$\left[{\mathrm{A}}_0\right]=$Initial concentration of A.

$\left[{\mathrm{A}}_{\mathrm{t}}\right]=$Concentration of A at time t.

$$\begin{gathered} \frac{{\mathrm{t}}_{99.9\%}}{{\mathrm{t}}_{50\%}}=\mathrm{x}=\frac{{\displaystyle \log \frac{\left[100\right]}{\left[100-99.9\right]}}}{{\displaystyle \log \frac{\left[100\right]}{\left[50\right]}}} \\ =\log \frac{{\displaystyle \frac{100}{0.1}}}{\log {\displaystyle }{\displaystyle 2}} \\ = \frac{\log {\displaystyle }{\displaystyle 1000}}{\log {\displaystyle }{\displaystyle 2}} \\ =\frac{3}{0.3}=10 \end{gathered}$$