Let $l$ bet he length of metal wire.
When wire is bent into a circular coil of radius r, then
$r=\frac{l}{2\pi }$
$\therefore Area,A=\pi {\left(\frac{l}{2\pi }\right)}^2=\pi \frac{l^2}{4{\pi }^2}$
$\therefore$ Magnetic dipole moment associated with circular coil,
${\mu }_c=iA=i\pi \left(\frac{l^2}{4{\pi }^2}\right)$
${\mu }_c=\frac{i{l}^2}{4\pi }$ … (i)
When metal wire is bent into a square coil then side of square
$a=\frac{l}{4}$
$\therefore Area,A=a^2=\frac{l^2}{16}$
$\therefore$ Magnetic dipole moment associated with square coil,
${\mu }_s=iA=i\frac{l^2}{16}$
$\therefore \frac{{\mu }_c}{{\mu }_s}=\frac{\frac{i{l}^2}{4\pi }}{\frac{i{l}^2}{16}}=\frac{16}{4\pi }=\frac{4}{\pi }$
Hence, the ratio of magnetic dipole moment of circular and square coil is $4:\pi$ .