Suppose the length of wire is $l$

Then, for the circle of n turns

The magnetic field, $B=\frac{{\mu }_{0}In}{2R}$ , where R is the radius of the circle

Then $B_s=\frac{{\mu }_0 I}{2R}$ =B......(i)

Here, $2\pi R=l$ .......(ii)

Now, if the same wire is turned into a coil of n turns of radius a then,

$2\pi a\times n=l$ .......(iii)

Thus, $2\pi R=2\pi a\times n$

$\Rightarrow a=\frac{R}{n}$

Now, the magnetic field for the coil of n turns, is

$B_n=\frac{{\mu }_{0}I}{2a}\times n$

$=\frac{{\mu }_{0}I}{2R}\times n^2$ $\left(\because a=\frac{R}{n}\right)$

$=n^{2}B$