Answer
$v_d = 1.14\times 10^{-4}~m/s$
Work Step by Step
We can find the number $N$ of copper atoms per $m^3$:
$N = \frac{9.0~g/cm^3}{64~g/mol}$
$N = 0.140625~mol/cm^3$
$N = (0.140625~mol/cm^3)(6.02\times 10^{23}~atoms/mol)(10^6~cm^3/m^3)$
$N = 8.4656\times 10^{28}~atoms/m^3$
We can find the number $n$ of conduction electrons per $m^3$:
$n = 1.3\times (8.4656\times 10^{28})$
$n = 1.1\times 10^{29}$
We can write an expression for the drift speed:
$v_d = \frac{I}{n~q~A}$
$I$ is the current
$n$ is the number of conduction electrons per $m^3$
$q$ is the charge of one electron
$A$ is the cross-sectional area
We can find the drift speed $v_d$:
$v_d = \frac{I}{n~q~A}$
$v_d = \frac{2.0~A}{(1.1\times 10^{29}~m^{-3})(1.6\times 10^{-19}~C)(1.00\times 10^{-6}~m^2)}$
$v_d = 1.14\times 10^{-4}~m/s$
