Answer
$2\;\rm A,\;5\times 10^{-5}\;m/s$
Work Step by Step
The current is the same in both wires according to the current conservation principle.
Hence,
$$I_1=I_2=I=\color{red}{\bf 2}\;\rm A$$
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,
$$v_{d1}=\dfrac{I}{A_1en_e}\tag 1$$
and
$$v_{d2}=\dfrac{I}{A_2en_e}\tag 2$$
Divide (2) by (1);
$$\dfrac{v_{d2}}{v_{d1}}=\dfrac{\dfrac{I}{A_2en_e}}{\dfrac{I}{A_1en_e}}=\dfrac{A_1}{A_2}$$
where $A=\pi r^2=\pi (D/2)^2=\pi D^2/4$ whereas $D$ is the diameter of the wire,
$$\dfrac{v_{d2}}{v_{d1}} =\dfrac{\dfrac{\pi D_1^2}{4}}{\dfrac{\pi D_2^2}{4}}=\dfrac{D_1^2}{D_2^2}$$
Thus,
$$v_{d2}=\dfrac{D_1^2}{D_2^2}v_{d1}$$
Plug the known;
$$v_{d2}=\dfrac{(1)^2}{(2)^2}(2\times 10^{-4})$$
$$v_{d2}=\color{red}{\bf 5\times 10^{-5}}\;\rm m/s$$
