Two cylindrical wires, A and B, are made of the same conducting...

The drift speed of electrons in a conductor is given by $v_d = \dfrac{I}{neA}$

Since both wires are made of the same material, the free electron density $n$ is the same for both wires. The cross-sectional area is $A = \pi r^2$.

Therefore, the drift speed is proportional to $\dfrac{I}{r^2}$

Given that $r_A = 2r_B$ and $I_A = 3I_B$

The ratio of drift speeds is $\dfrac{v_{dA}}{v_{dB}} = \left(\dfrac{I_A}{I_B}\right) \left(\dfrac{r_B}{r_A}\right)^2$

Substituting the given values:

$\dfrac{v_{dA}}{v_{dB}} = (3) \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4}$

The ratio of the drift speed of electrons in wire A to that in wire B is $3 : 4$.

Answer: $3 : 4$