Time required for completion of 99 . 9 % of first order reaction is...

For first order reaction:
$kt = log\dfrac{\left[{A}_{0}\right]}{\left[{A}_{t}\right]}$

\left[{A}_{0}\right] =$Initial concentration of A.</p><p>\left[{A}_{t}\right] =$Concentration of A at time t.

\dfrac{{t}_{99 . 9 %}}{{t}_{50 %}} = x = \dfrac{log\dfrac{\left[100\right]}{\left[100 - 99 . 9\right]}}{log\dfrac{\left[100\right]}{\left[50\right]}} = log\dfrac{\dfrac{100}{0 . 1}}{\log 2} = \dfrac{\log 1000}{\log 2} = \dfrac{3}{0 . 3} = 10