In a metal wire the electrons have a relaxation time of 1.6 × 10⁻¹⁴ s…

Step 1. The drift speed. $v_d = \dfrac{eE\tau}{m}$. The top is $1.6 \times 10^{-19} \times 0.12 \times 1.6 \times 10^{-14} = 3.07 \times 10^{-34}$. So $v_d = \dfrac{3.07 \times 10^{-34}}{9.1 \times 10^{-31}\ \text{kg}} = 3.38 \times 10^{-4}\ \text{m/s}$, that is $0.338$ mm/s.

Step 2. The mobility. $\mu = \dfrac{v_d}{E} = \dfrac{3.38 \times 10^{-4}\ \text{m/s}}{0.12\ \text{V/m}} = 2.81 \times 10^{-3}\ \text{m}^2/(\text{V s})$.