Answer
$5.56\times 10^{-6 }\;\rm m/s$
Work Step by Step
The current is the same in both wires according to the current conservation principle.
Hence,
$$I_1=I_2=I\tag 1$$
where $I_1$ is at the 3-mm diameter wire and $I_2$ is at the 1-mm wire.
However, the current density is not the same in both cases.
Recalling that the drift speed is given by
$$v_d=\dfrac{J}{en_e}$$
where $J=I/A$,
$$v_d=\dfrac{I}{Aen_e}$$
So that,
$$v_{d1}=\dfrac{I}{A_1en_e}\tag 2$$
and
$$v_{d2}=\dfrac{I}{A_2en_e}\tag 3$$
Divide (2) by (3);
$$\dfrac{v_{d1}}{v_{d2}}=\dfrac{\dfrac{I}{A_1en_e}}{\dfrac{I}{A_2en_e}}=\dfrac{A_2}{A_1}$$
where $A=\pi r^2=\pi (D/2)^2=\pi D^2/4$ whereas $D$ is the diameter of the wire,
$$\dfrac{v_{d1}} {v_{d2}}=\dfrac{\dfrac{\pi D_2^2}{4}}{\dfrac{\pi D_1^2}{4}}=\dfrac{D_2^2}{D_1^2}$$
Thus,
$$v_{d1}=\dfrac{D_2^2}{D_1^2}v_{d2}$$
Plug the known;
$$v_{d1}=\dfrac{(1)^2}{(3)^2}(0.5\times 10^{-4})$$
$$v_{d1}=\color{red}{\bf 5.56\times 10^{-6}}\;\rm m/s$$
