Answer
We can rank the wires in order of decreasing drift velocity:
$b \gt d \gt a = e \gt f \gt c$
Work Step by Step
We can write an expression for the drift velocity:
$v_d = \frac{I}{n~q~A} = \frac{I}{n~q~\pi~r^2}$
$I$ is the current
$n$ is the number of electrons per unit of volume
$q$ is the charge of one electron
$A$ is the cross-sectional area
$r$ is the cross-sectional radius
We can find an expression for the drift velocity in each case:
(a) $v_d = \frac{80\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{80\times 10^3}{n~q~\pi}$
(b) $v_d = \frac{80\times 10^{-3}}{n~q~\pi~(0.5\times 10^{-3})^2} = \frac{320\times 10^3}{n~q~\pi}$
(c) $v_d = \frac{40\times 10^{-3}}{n~q~\pi~(2\times 10^{-3})^2} = \frac{10\times 10^3}{n~q~\pi}$
(d) $v_d = \frac{160\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{160\times 10^3}{n~q~\pi}$
(e) $v_d = \frac{20\times 10^{-3}}{n~q~\pi~(0.5\times 10^{-3})^2} = \frac{80\times 10^3}{n~q~\pi}$
(f) $v_d = \frac{40\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{40\times 10^3}{n~q~\pi}$
We can rank the wires in order of decreasing drift velocity:
$b \gt d \gt a = e \gt f \gt c$
