Answer
$2.4\;\rm A$
Work Step by Step
The current in the wire is given by the current density
$$J=\dfrac{I}{A}=\sigma E$$
Hence,
$$I=A\sigma E$$
$$I=\pi r^2\sigma_{aluminum} E\tag 1$$
Now we need to find the electric field.
We know that the electric field of a charged ring on its axis is given by
$$E_x=\dfrac{1}{4\pi \epsilon_0}\dfrac{Qx}{(x^2+R^2)^{3/2}}$$
And since the electron moves from one ring to the other, we can assume that the left one is negatively charged while the other is positively charged.
This means that the net electric field at the center point between the two rings from the left-negatively charged ring is toward the left while the net electric field at the same point from the right-positively charged ring is toward the left too.
Hence, the net electric field magnitude is the sum of both and its direction is also to the left.
$$E_x=\dfrac{1}{4\pi \epsilon_0}\dfrac{2Qx}{(x^2+R^2)^{3/2}}$$
Plug that into (1),
$$I=\pi r^2\sigma_{aluminum} \dfrac{1}{4\pi \epsilon_0}\dfrac{2Qx}{(x^2+R^2)^{3/2}}$$
where $Q=Ne$ where $N$ is the number of the electrons.
$$I=\pi r^2\sigma_{aluminum} \dfrac{1}{4\pi \epsilon_0}\dfrac{2Nex}{(x^2+R^2)^{3/2}}$$
Now we need to plug the known;
$$I=\pi (1.5\times 10^{-3})^2(3.5\times 10^7)(9\times 10^9)(\dfrac{2(20)(1.6\times 10^{-19})(1\times 10^{-3})}{([1\times 10^{-3}]^2+[1.5\times 10^{-3}]^2)^{3/2}}$$
$$I=\color{red}{\bf 2.4}\;\rm A$$
