Two uniform wires of same material are vibrating under the same tension. If the $1^{\text {st }}$ overtone of $1^{\text {st }}$ wire is equal to the $2^{\text {nd }}$ overtone of $2^{\text {nd }}$ wire and radius of $1^{\text {st }}$ wire is twice the radius of $2^{\text {nd }}$ wire, the ratio of length of $1^{\text {st }}$ wire to that $2^{\text {nd }}$ wire is

The fundamental frequency relationship for a string fixed at both ends gives the $n$th overtone frequency as $f_n = (n+1)\frac{v}{2L}$, where $v = \sqrt{\frac{T}{\mu}}$ is the wave speed and $\mu = \rho\pi r^2$ is the mass per unit length for a wire of radius $r$ and density $\rho$.

Substituting yields $f_n = \frac{(n+1)}{2Lr}\sqrt{\frac{T}{\rho\pi}}$. For the first wire's first overtone ($n=1$), $f_1 = \frac{1}{L_1r_1}\sqrt{\frac{T}{\rho\pi}}$, and for the second wire's second overtone ($n=2$), $f_2 = \frac{3}{2L_2r_2}\sqrt{\frac{T}{\rho\pi}}$.

Setting $f_1 = f_2$ and canceling the common factor $\sqrt{\frac{T}{\rho\pi}}$ gives $\frac{1}{L_1r_1} = \frac{3}{2L_2r_2}$. Given $r_1 = 2r_2$, substitution yields $\frac{1}{2L_1r_2} = \frac{3}{2L_2r_2}$, which simplifies to $\frac{1}{L_1} = \frac{3}{L_2}$.

The ratio is therefore $L_1:L_2 = 1:3$.

Final answer: $\boxed{\text{A}}$