A long wire carrying a steady current is bent into a circle of...

Let the length of the wire be $L$.

For a single turn circular loop of radius $R$, the length is $L = 2\pi R \Rightarrow R = \dfrac{L}{2\pi}$.

The magnetic field at the centre is $B = \dfrac{\mu_0 I}{2R}$.

When the same wire is bent into a circular loop of $n$ turns of radius $r$, the length is $L = n(2\pi r) \Rightarrow r = \dfrac{L}{2\pi n} = \dfrac{R}{n}$.

The new magnetic field at the centre is $B' = \dfrac{\mu_0 n I}{2r}$.

Substituting $r = \dfrac{R}{n}$, we get $B' = \dfrac{\mu_0 n I}{2(R/n)} = n^2 \left(\dfrac{\mu_0 I}{2R}\right) = n^2 B$.

Answer: $\text{n}^2\text{B}$