Answer
$(i_e)_d < (i_e)_b < (i_e)_e < (i_e)_a = (i_e)_c$
Work Step by Step
Recall that $I = A\sigma \vec{E} =$ and $I = ei_e$, so therefore $\displaystyle \frac{A\sigma \vec{E}}{e} = i_e$. This means that $i_e$ is directly related/proportional to $\vec{E}$ ($i_e \propto \vec{E}$). We can now rank the charged rings from most electric field to least. For the sake of ranking, let a ring with 2 positive marks be of charge $+q$, and a ring of 2 negative marks be of charge $-q$.
a.) Since both rings have the same charge (positive) of the same magnitude (each have a charge of $+q$), then the net electric field cancels out and $\vec{E} = 0$
b.) Since the rings have the opposite charges, the electric field strength points right for both rings so the electric field strength is $2\vec{E}$.
c.) Again, both rings have the same charge (both have $+2q$ charge), so the net electric field is 0.
d.) Since both rings have opposite charges double the magnitude of those in b.), the electric field is $4\vec{E}$.
e.) Although both rings have the same type of charge, their magnitudes are different. This means that we have to negate the electric fields from each other so $2\vec{E} - \vec{E} = \vec{E}$
From largest to smallest, the electron currents $i_e$ rank:
$(i_e)_d < (i_e)_b < (i_e)_e < (i_e)_a = (i_e)_c$
