A copper wire is made of two pieces joined end to end, as the figure shows…

(a) Thick piece. From $I = n A e v_d$, $v_d = \dfrac{I}{n A e}$. First, $n A e = 8.5 \times 10^{28} \times 3.142 \times 10^{-6} \times 1.6 \times 10^{-19} = 4.27 \times 10^{4}\ \text{C/m}$. Then $v_d = \dfrac{3.0\ \text{A}}{4.27 \times 10^{4}\ \text{C/m}} = 0.0702\ \text{mm/s}$.

(b) Thin piece. The area is four times smaller, so the drift speed is four times larger: $4 \times 0.0702\ \text{mm/s} = 0.2809\ \text{mm/s}$.

(c) The field. The current density in the thin piece is $J = \dfrac{I}{A_2} = 3.820 \times 10^{6}\ \text{A/m}^2$. Then $E = \rho J = (1.7 \times 10^{-8}\ \Omega\,\text{m})(3.820 \times 10^{6}\ \text{A/m}^2) = 0.0649\ \text{V/m}$, that is $64.9\ \text{mV/m}$.

(d) The potential difference. The field is uniform in the thin piece, so $\Delta V = E L = 0.065\ \text{V/m} \times 0.50\ \text{m} = 0.0325\ \text{V}$, that is $32.5\ \text{mV}$.

(e) Relaxation time. In a metal, $\rho = \dfrac{m}{n e^2 \tau}$, so $\tau = \dfrac{m}{n e^2 \rho}$. First, $n e^2 \rho = 8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 1.7 \times 10^{-8} = 3.70 \times 10^{-17}$ in SI units. Then $\tau = \dfrac{9.1 \times 10^{-31}}{3.70 \times 10^{-17}}\ \text{s} = 24.6 \times 10^{-15}\ \text{s}$, which is $24.6\ \text{fs}$.