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**Hint:**Note that, the problem is of resistance including the length and area of the wire. The relation between conductivity and resistance should be used here. Next, convert the volume into length.

Here given that the new wire is made of $n$-fold. The length of the new wire has to be calculated using the given fold number. And then the resistance is to be calculated using the new length of the wire.

**Formula used:**The length of the wire is $l$ and the area of the wire is $A$ , so the resistance of the wire is $R = \rho \dfrac{l}{A}$

$\rho $ is the conductivity.

The area can be written as, $A = \dfrac{V}{l}$

The new length of the wire after stretching, $L = nl$

**Complete step-by-step solution:**

The resistance of the wire of length $l$ and area $A$, before stretching

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

$\rho $ is the conductivity.

As the volume will remain unchanged, the equation should be converted in terms of volume.

So, the area can be written as, $A = \dfrac{V}{l}$

$\therefore R = \rho \dfrac{l}{{\dfrac{V}{l}}}$

$ \Rightarrow R = \dfrac{{\rho {l^2}}}{V}.............(1)$

The wire is elongated $n$-fold to make a new uniform wire.

So, The new length of the wire after stretching, $L = nl$

Now the resistance will be,

$ \Rightarrow {R'} = \dfrac{{\rho {L^2}}}{V}$

$ \Rightarrow {R'} = \dfrac{{\rho {{(nl)}^2}}}{V}$

$ \Rightarrow {R'} = \dfrac{{\rho {n^2}{l^2}}}{V}$

$ \Rightarrow {R'} = {n^2}\dfrac{{\rho {l^2}}}{V}$

$ \Rightarrow {R'} = {n^2}R$ [from the eq. (1)]

**Hence, the right answer is in option (b).**

**Note:**The resistance of a circuit element or device is outlined because of the magnitude ratio of the voltage applied to the electrical current that flows through it:

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

If the resistance is constant over a substantial vary of voltage, then Ohm's law, \[I{\text{ }} = {\text{ }}\dfrac{V}{R}\] , may be accustomed to predict the behavior of the material. Though the definition involves DC and voltage, an identical definition holds for the AC application of resistors.

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