1:1
1:9
9:1
1:3
Magnetic field due to the first loop ,B1=μ₀I/2r
For 2nd wire, there are three loops (n=3) so total current enclosed by the loops ,Ien=I/3+I/3+I/3=I
Radius of second loop=r/3 because the original wires are identical.
magnetic field B2=μ₀Ien/2(r/3) = μ₀3I/2r
B1/B2=1/3