If a single-turn circular coil ( of radius R_1 ) carrying a current...

Let the total length of the wire be $L$.

For the single-turn coil of radius $R_1$, the length is $L = 2\pi R_1$.

For the $N$-turn coil of radius $R_2$, the length is $L = N(2\pi R_2)$.

Equating the lengths, $2\pi R_1 = N(2\pi R_2) \Rightarrow R_2 = \dfrac{R_1}{N}$.

The magnetic field at the centre of the original coil is $B_1 = \dfrac{\mu_0 I}{2 R_1}$.

The magnetic field at the centre of the new coil is $B_2 = \dfrac{\mu_0 N I}{2 R_2}$.

Substituting $R_2 = \dfrac{R_1}{N}$ into the expression for $B_2$:

$B_2 = \dfrac{\mu_0 N I}{2 \left(\dfrac{R_1}{N}\right)} = N^2 \left(\dfrac{\mu_0 I}{2 R_1}\right) = N^2 B_1$.

The ratio of the magnetic field at the centre of the new coil to that of the original coil is $\dfrac{B_2}{B_1} = N^2$.

Answer: $N^2$