Answer
$i_{e,3} > i_{e,2} > i_{e,1} > i_{e,3}$
Work Step by Step
Recall $i_e = n_eA_{\perp}v_d$ and $A_{\perp} = \pi r^2$
Since all of the wires are made of the same material, we don't need to worry about $n_e$. That is, $i_e \propto A_{\perp}v_d = \pi r^2v_d$
The electron current for each wire is:
$i_{e,1} = \pi (r)^2(v_d) = \pi r^2 v_d$
$i_{e,2} = \pi (r)^2(2v_d) = 2\pi r^2 v_d$
$i_{e,3} = \pi (2r)^2(v_d) = 4\pi r^2 v_d$
$\displaystyle i_{e,4} = \pi (\frac{1}{2}r)^2(2v_d) = \frac{1}{2}\pi r^2 v_d$
So in ranking from largest to smallest, the order is $i_{e,3} > i_{e,2} > i_{e,1} > i_{e,3}$.
