20 % first order reaction is completed in 50 minutes. Time required for the completion of 60 % of the reaction is min.

 K_20%  =K_60%

$\frac{2.303}{50}\log \frac{100}{80}=\frac{2.303}{\mathrm{t}}\log \frac{100}{40}$

$\frac{\log 10-\log 8}{50} = \frac{\log 10-\log 4}{\mathrm{t}}$

$\frac{1-0.9030}{50} = \frac{1-0.6020}{\mathrm{t}}$

$\frac{0.097}{50} = \frac{0.398}{\mathrm{t}}$

   $$\begin{gathered} \mathrm{t} =\frac{0.398\times 50}{0.097} \\ \mathrm{t}=205 \end{gathered}$$