Answer
$v_d = 1.8\times 10^{-15}~m/s$
Work Step by Step
In part (a), we found that $J = 2.4\times 10^{-5}~A/m^2$
We can find the electron drift speed:
$J = n~e~v_d$
$v_d = \frac{J}{n~e}$
$v_d = \frac{2.4\times 10^{-5}~A/m^2}{(8.49\times 10^{28}~m^{-3})(1.6\times 10^{-19}~C)}$
$v_d = 1.8\times 10^{-15}~m/s$
