In the above case the volume and specific resistivity of the wire does not change, it means,
 $$\begin{gathered} V_1=V_2\Rightarrow \pi {r_1}^2{l}_1=\pi {r_2}^2{l}_2 \\ \Rightarrow \frac{{r_1}^2}{{r_2}^2}=\frac{l_2}{l_1} \end{gathered}$$

Now recall the formula of resistance in terms of specific resistivity, 
$R=\rho \times \frac{l}{\pi r^2}$
$\Rightarrow \frac{R_1}{R_2}=\frac{\frac{l_1}{\pi {r_1}^2}}{{\displaystyle \frac{l_2}{\pi {r_1}^2}}}={\left(\frac{l_1}{l_2}\right)}^2$

$$\begin{gathered} \Rightarrow {\left(\frac{l_1}{l_2}\right)}^2=\left(\frac{R_1}{R_2}\right)\Rightarrow \frac{2∆\left(\frac{l_1}{l_2}\right)}{\left(\frac{l_1}{l_2}\right)}\times 100=\frac{∆\left(\frac{R_1}{R_2}\right)}{\left(\frac{R_1}{R_2}\right)}\times 100=4\% \\ \Rightarrow \frac{∆\left(\frac{l_1}{l_2}\right)}{\left(\frac{l_1}{l_2}\right)}\times 100=2\% \end{gathered}$$