Answer
$v_d = 5.87\times 10^{-5}~m/s$
Work Step by Step
We can write an expression for the drift speed:
$v_d = \frac{I}{n~q~A} = \frac{I}{n~q~\pi~r^2}$
$I$ 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 the drift speed $v_d$:
$v_d = \frac{I}{n~q~\pi~r^2}$
$v_d = \frac{2.50~A}{(8.47\times 10^{28}~m^{-3})(1.6\times 10^{-19}~C)~(\pi)~(1.00\times 10^{-3}~m)^2}$
$v_d = 5.87\times 10^{-5}~m/s$
