$\because$ 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 $1:2$