As per the given question, the change in the length is $∆l=2l$.

Resistance of a wire is given by, $R=\frac{\rho l}{A}$

Therefore, $R_1=\rho \frac{l_1}{ A_1}$

When the length is starched by $2l_1$, final length will be $l_2=l_1+2l_1=3l_1$. As the volume remains constant $l_2{A}_2=l_1{A}_1 \Rightarrow A_2=\frac{A_1}{3}$. Therefore,

$R_2=\rho \left(\frac{3l_1}{{\displaystyle \frac{ A_1}{3}}}\right)=9\rho \frac{l_1}{ A_1}$

$\therefore \frac{R_2}{R_1}=9$