Given:

The ratio of lengths: $\frac{L_A}{L_B} = \frac{1}{3}$

The ratio of diameters: $\frac{d_A}{d_B} = 2$

Both wires are subject to the same stretching force, and since they are made of the same material, their Young's modulus ($Y$) is the same.

The elongation ($\Delta L$) of a wire subject to a force is given by:

$ \Delta L = \frac{F \cdot L}{A \cdot Y} $

where $F$ is the force applied, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.

For wires $A$ and $B$:

$\Delta L_A = \frac{F_A \cdot L_A}{A_A \cdot Y_A}$

$\Delta L_B = \frac{F_B \cdot L_B}{A_B \cdot Y_B}$

Since $F_A = F_B$ and $Y_A = Y_B$, the ratio of their elongations becomes:

$ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} $

The cross-sectional area $A$ is a function of diameter, $A = \frac{\pi}{4} d^2$. Therefore,

$ A_A = \frac{\pi}{4} d_A^2 \quad \text{and} \quad A_B = \frac{\pi}{4} d_B^2 $

Substituting these into the elongation ratio:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{\frac{\pi}{4} d_B^2}{\frac{\pi}{4} d_A^2}\right) = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2 $

Substitute the given ratios:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2 = \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = \frac{1}{12} $

Thus, the ratio of the elongations of wires $A$ and $B$ is $\frac{1}{12}$.