$6.25\%$ of the initial value means, final number of nuclei should be, $\left(\frac{N_0}{16}\right)$.

Number of nuclei remaining after $n$ half lives is $N=\frac{N_0}{2^n}$
$6.25\% N_0=\frac{N_0}{2^n}$
$\frac{6.25}{100}{N}_0=\frac{N_0}{2^n} \Rightarrow \frac{625}{{10}^4}=\frac{1}{2^n}$
$\Rightarrow 2^n=\frac{{10}^4}{5^4} \Rightarrow 2^n=2^4$
$\Rightarrow n=4$

Therefore, time taken will be, $n\times t_{\frac{1}{2}}=4\times 5=20 years$