Let \(l\) be the length of the wire, then
\(B_1=\frac{\mu_0}{2 \pi} \cdot \frac{1 \times I}{l / 2 \pi}=\frac{\mu_0 I}{l}\)
and \(B_2=\frac{\mu_0}{2 \pi} \cdot \frac{2 \times I}{l / 4 \pi}=\frac{4 \mu_0 I}{l}\)
Therefore, \(\frac{B_1}{B_2}=\frac{1}{4}\)
or, \(B_1: B_2=1: 4\)