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**Hint:**Use the formula of the self-induction of the solenoid given below, substitute the formula of the length of the wire and the area of the wire in the above formula and simplify it to obtain the relation for the self-induction of the solenoid.

**Formula used:**

The self-induction is given by

$L = \dfrac{{{\mu _0}{N^2}A}}{l}$

Where $L$ is the self-induction of the solenoid, ${\mu _0}$ is the magnetic permeability, $l$ is the length of the solenoid and $A$ is the area of each turn in the solenoid.

**Complete step by step solution:**

Let us consider the wire is of length $x$

It is known that the length of the solenoid is $2\pi rN$ . Since the wire is in the shape of the cylinder, the cross sectional area is $A = \pi {r^2}$.

Use the formula of the self-induction,

$L = \dfrac{{{\mu _0}{N^2}A}}{l}$

Substitute the formula of $N = \dfrac{x}{{2\pi r}}$ and the area as $\pi {r^2}$ in the above formula.

$L = \dfrac{{{\mu _0}{{\left( {\dfrac{x}{{2\pi r}}} \right)}^2}\left( {\pi {r^2}} \right)}}{l}$

By simplifying the above equation, we get

$L = \dfrac{{{\mu _0}\left( {\dfrac{{{x^2}}}{{4{\pi ^2}{r^2}}}} \right) \times \left( {\pi {r^2}} \right)}}{l}$

By canceling the similar terms in the above step.

$L = \dfrac{{{\mu _0}\left( {\dfrac{{{x^2}}}{{4\pi }}} \right)}}{l}$

By bringing the length of the wire in the left side and other terms in the right side of the equation.

$\dfrac{{\,Ll}}{{{\mu _0}}} = \left( {\dfrac{{{x^2}}}{{4\pi }}} \right)$

By the further simplification of the above equation,

$x = \sqrt {\dfrac{{4\pi Ll}}{{{\mu _0}}}} $

Hence the length of the wire obtained is $\sqrt {\dfrac{{4\pi Ll}}{{{\mu _0}}}} $ .

**Thus the option (C) is correct.**

**Note:**The wire is in the form of the slender cylinder, hence the cross sectional area is considered as the $\pi {r^2}$ . The number of the rotation is calculated by dividing the whole length of the wire by the circumferential area of the wire as $2\pi r$.

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