What is $n$ ?

We know,

$R=\rho \frac{l}{A}$

Here, $R$ is the resistance of the conductor, $l$ is the length of the conductor and $A$ is the area of cross section of the conductor.

As the conductor is stretched, its area of cross section would decrease but the volume would remain same.

$\therefore R=\rho \frac{l}{A}\cdot \frac{l}{l} = \rho \frac{l^2}{V}$

Here, $V$ is the volume of the conductor.

$\therefore R \propto l^2$

As $l$ is increased by $3$ times, $R$ would be $\boxed{9}$ times the original resistance.

As the material is not changed, the specific resistivity is not changed, so for a conductor of resistance $R$ , area of cross-section $a$ and length $l$ . When conductor is stretched $3$ times, the length increases to $3l$ and area of cross section decreases to $\dfrac {a}{3}$

$\text {CASE 1}:-$
$\begin{aligned} & \rho & = \dfrac {R×a}{l} \end{aligned}$

$\text {CASE 2}:-$
$\begin{aligned} & \rho & = \dfrac {nR×\dfrac {a}{3}}{3l} \end{aligned}$

$\text {Combining both the equations}$ :-
$\begin{aligned} & \dfrac {R×a}{l} = \dfrac {nR×\dfrac {a}{3}}{3l}\\ & \dfrac {R}{nR} = \dfrac {a × l}{3 × 3l × a}\\ & \require {cancel}{\dfrac {\cancel R}{n \cancel R}} = {\dfrac {\cancel a × \cancel l}{3 × 3\cancel l × \cancel a}}\\ & n = \boxed {9} \end{aligned}$

As the length becomes three times the initial length , the area of cross section becomes one-third the initial area of cross section . As we know R = Resistivity * Length / Area of Cross section
It will be 9 times the initial resistance .