Given that the half-life of a radioactive substance is $20\, min .$ So, $t_{1 / 2}=20\, min$.
For $20 \%$ decay, we have $80 \%$ of the substance left, hence
$\frac{80\, N_{0}}{100}=N_{0} e^{-\lambda t_{20}}\,\,\,...(i)$
where $N_{0}=$ initial undecayed substance and $t_{20}$ is the time taken for $20 \%$ decay.
For $80 \%$ decay, we have $20 \%$ of the substance left, hence
$\frac{20 N_{0}}{100}=N_{0} e^{-\lambda t_{80}}\,\,\,...(ii)$
Dividing Eq. (i) and Eq. (ii), we get
$4 =e^{\lambda\left(t_{80}-t_{20}\right)} $
$\Rightarrow \ln 4 =\lambda\left(t_{80}-t_{20}\right)$
(taking log on both sides)
$\Rightarrow 2 \ln 2=\frac{0.693}{t_{1 / 2}}\left(t_{80}-t_{20}\right)$
$\Rightarrow t_{80}-t_{20}=2 \times t_{1 / 2}=40\, min$