b
Let mass per unit length of wires are \(\mu_{1}\) and \(\mu_{2}\) respectively.

\(\because\) Materials are same, so density \(\rho\) is same.

\(\mu_{1}=\frac{\rho \pi r^{2} L}{L}=\mu \text { and } \mu_{2}=\frac{\rho 4 \pi r^{2} L}{L}=4 \mu\)

Tension in both are same \(=\mathrm{T},\) let speed of wave in wires are \(V_{1}\) and \(V_{2}\)

\(V_{1}=\frac{V_{1}}{2 L}=\frac{V}{2 L} \& f_{02}=\frac{V_{2}}{2 L}=\frac{V}{4 L}\)

Frequency at which both resonate is \(L.C.M.\) of both frequencies i.e. \(\frac{\mathrm{V}}{2 \mathrm{L}}\)

Hence number of loops in wires are \(1\) and \(2\) respectively

So, ratio of number of antinodes is \(1:2\)