Explanation :
Head loss due to frriction $\style{font-family:'Times New Roman'}{\mathrm{hf}=\frac{4\mathrm{flv}^2}{2\mathrm g.\mathrm d}=\frac{4\mathrm{fl}(\mathrm Q/\mathrm A)^2}{2\mathrm g.\mathrm d}}$
$\style{font-family:'Times New Roman'}{\begin{array}{l}\mathrm A=\mathrm{πr}^2\\\;\;\;=\mathrm\pi\left(\frac{\mathrm d}2\right)^2=\frac{\mathrm{πd}^2}4\end{array}}$
So $\style{font-family:'Times New Roman'}{hf=\frac{4f.l.\left(Q.4/\mathrm{πd}^2\right)^2}{2.g.d}=\frac{4f.l.Q^2.16}{\mathrm\pi^2.\mathrm d^4.2.\mathrm g.\mathrm d}}$
$\style{font-family:'Times New Roman'}{\therefore\mathrm{hf}\;\propto\frac{\mathrm l}{\mathrm d^5}\left(\because\mathrm F,\mathrm Q,\mathrm\pi,\mathrm g\;\mathrm{are}\;\mathrm{const}.\right)}$
$\style{font-family:'Times New Roman'}{\begin{array}{l}{\mathrm{hf}}_1\propto\frac{\mathrm l}{{\mathrm d}_1^5}\;\;\;\;\;\;\;\;\;\\\;{\mathrm{hf}}_2\propto\frac{\mathrm l^2}{{\mathrm d}_2^5}=\frac{2.{\mathrm l}_1}{\left(\mathrm d1/2\right)^5}=\frac{{\mathrm l}_1}{{\mathrm d}_1^2}.64\end{array}}$
$\begin{array}{l}\frac{{\mathrm{hf}}_1}{{\mathrm{hf}}_2}=\frac1{64}\\{\mathrm{hf}}_2=64\;{\mathrm{hf}}_1\end{array}$