Silver has 5.86 × 10²⁸ free electrons in each cubic metre, and a resistivity of 1.6 × 10⁻⁸ Ω m…

Step 1. The relaxation time. $\tau = \dfrac{m}{ne^2\rho}$. The bottom is $5.86 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 1.6 \times 10^{-8} = 2.40 \times 10^{-17}$ in SI units. So $\tau = \dfrac{9.1 \times 10^{-31}}{2.40 \times 10^{-17}} = 3.79 \times 10^{-14}\ \text{s}$.

Step 2. The distance. $\lambda = v\tau = 1.0 \times 10^6\ \text{m/s} \times 3.79 \times 10^{-14}\ \text{s} = 3.79 \times 10^{-8}\ \text{m}$, that is $37.9$ nm.