If $10\%$ of a radioactive material decay in 5 days, then the amount of original material left after 20 days is approximately:

$N=N_0{e}^{-\lambda t}$

When $10\%$decay, $N=90\%$

$\Rightarrow 0.9N_0=N_0{e}^{-5\lambda }$

$\Rightarrow \log \left(0.9\right)=-5\lambda$ .....(i)

Let assume after 20 days fraction in decayed be x

$x{\mathrm{N}}_0={\mathrm{N}}_0{e}^{-20\lambda }$

$\log x=-20\lambda$ ..........(ii)

from eqn (i) & (ii)

$\frac{1}{4}=\frac{\log \left(0\cdot 9\right)}{\log x}$

$\Rightarrow \log x=4\log \left(0.9\right)$

$\Rightarrow x=(0.9{)}^4=0.65 =65\%$