A long conducting wire having a current I flowing through it, is...

As in both cases, length of the wire will remain the same, therefore

$l = \left(2\piR\right)N = \left(2\pir\right)n ⇒ \dfrac{R}{r} = \dfrac{n}{N}$

Now magnetic field due to single circular coil is given by, $B = \left(\dfrac{{\mu}_{0}I}{2r}\right)$.

Therefore, required ratio of total magnetic field wil be,

$\dfrac{{B}_{1}}{{B}_{2}} = \dfrac{N\left(\dfrac{{\mu}_{0}I}{2R}\right)}{n\left(\dfrac{{\mu}_{0}I}{2r}\right)} = \left(\dfrac{{N}^{2}}{{n}^{2}}\right)$