Then, the lenght of wire which is stretched $=a l$ $R_{l}=R_{(w i r e)}=\frac{\rho l}{A}$

$R_{l-a l}=\frac{\rho(l-a l)}{A}=R_{l(1-a)}$

$R_{a l}=\frac{\rho a l}{A}=a R_{l}$

On stretching $'al'$ length of wire, new length of $^{\prime} a l^{\prime}$ part$:$

$l-a l+l^{\prime}=\frac{3 l}{2}$

$l^{\prime}=l\left(\frac{1}{2}+a\right)$

Volume of wire $' a l^{\prime}$ is same$:$

$l^{\prime} A^{\prime}=(a l) A$

$l\left(\frac{1}{2}+a\right) A^{\prime}=(a l) A$

$A^{\prime}=\frac{a A}{\left(\frac{1}{2}+a\right)}$

Now, resistance of stretched parts,

$R^{\prime}=\frac{\rho \times l^{\prime}}{A^{\prime}}=\frac{\rho \times l\left(\frac{1}{2}+a\right)}{\frac{a A}{\frac{1}{2}+a}}=\frac{\rho l}{A} \frac{\left(\frac{1}{2}+a\right)^{2}}{a}$

$\Rightarrow R^{\prime}=\frac{R_{l}\left(\frac{1}{2}+a\right)^{2}}{a}$

According to the question,

$R_{l-a l}+R^{\prime}=4 R_{l}$

$\Rightarrow R_{l(1-a)}+R_{l} \frac{\left(\frac{1}{2}+a\right)^{2}}{a}=4 R_{l}$

$\Rightarrow a-a^{2}+\frac{1}{4}+a^{2}+a=4 a$

$\Rightarrow \frac{1}{4}+2 a=4 a$

$\Rightarrow \frac{1}{4}=4 a-2 a$

$\Rightarrow \frac{1}{4}=2 a$

$\Rightarrow a=\frac{1}{8}$