A long wire carrying a steady current is bent into a circular loop...

$B = \dfrac{{\mu}_{0}I}{2r}$.

When $n$ turns are made, radius becomes ${r}^{′}$.

$n \times 2\pi{r}^{′} = 2\pir ⇒ {r}^{′} = \dfrac{r}{n}$.

Now, ${B}^{′} = \dfrac{{\mu}_{0}nI}{2{r}^{′}} = {n}^{2}\dfrac{{\mu}_{0}I}{2r} = {n}^{2}B$.