A copper wire is 1.8 mm across and carries 6.0 A…

Step 1. The area. $A = \dfrac{\pi d^2}{4} = \dfrac{\pi \times (1.8 \times 10^{-3}\ \text{m})^2}{4} = 2.55 \times 10^{-6}\ \text{m}^2$.

Step 2. The drift speed. $v_d = \dfrac{I}{nAe}$. The bottom is $8.5 \times 10^{28} \times 2.55 \times 10^{-6} \times 1.6 \times 10^{-19} = 3.46 \times 10^4\ \text{C/m}$. So $v_d = \dfrac{6.0\ \text{A}}{3.46 \times 10^4\ \text{C/m}} = 1.73 \times 10^{-4}\ \text{m/s}$, that is $0.173$ mm/s.

Step 3. The time. $t = \dfrac{2.0\ \text{m}}{1.73 \times 10^{-4}\ \text{m/s}} = 1.15 \times 10^4\ \text{s}$, that is $3.2$ hours.