Solution

$\textbf{a.}$ In the classical electron theory of conductivity, the motion of an electron between collisions with lattice ions is considered uniformly accelerated under the action of an external electric field $\vec{E}$. The acceleration of the electron is:
$$\vec{a} = -\frac{e\vec{E}}{m_e}$$
where $e$ is the elementary charge, and $m_e$ is the mass of the electron.

Since immediately after a collision any direction of velocity is equally probable, the average velocity of the random motion of electrons immediately after a collision is zero: $\langle \vec{v}_0 \rangle = 0$.

At any given time $t$ after the last collision, the average velocity of directed motion (drift velocity) $\upsilon_d$ is determined by averaging over all free flight times. With strict consideration of the flight time distribution (exponential law), the average value of the acquired velocity is:
$$\upsilon_d = a\tau = \frac{eE\tau}{m_e}$$

The electric current density $j$ is related to the drift velocity and the electron concentration $n_e$ by the relation:
$$j = en_e\upsilon_d$$

Substituting the expression for the drift velocity, we obtain:
$$j = \frac{e^2 n_e \tau}{m_e} E$$

According to Ohm's law in differential form, the current density is proportional to the electric field strength:
$$j = \lambda E$$

Comparing the two expressions, we find the required specific conductivity $\lambda$:
$$\lambda = \frac{e^2 n_e \tau}{m_e}$$

$\textbf{b.}$ To estimate the average time between collisions $\tau$ for copper, we express it from the obtained formula:
$$\tau = \frac{\lambda m_e}{e^2 n_e}$$

From reference data, the specific conductivity of copper is known: $\lambda \approx 5.9 \cdot 10^7 \, \text{S/m}$.
Let us estimate the concentration of conduction electrons $n_e$, assuming that there is one free electron per copper atom in the crystal lattice:
$$n_e = \frac{\rho_{Cu} N_A}{M}$$
where $\rho_{Cu} \approx 8900 \, \text{kg/m}^3$ is the density of copper, $M \approx 0.0635 \, \text{kg/mol}$ is the molar mass, and $N_A \approx 6.02 \cdot 10^{23} \, \text{mol}^{-1}$ is Avogadro's number.
$$n_e = \frac{8900 \cdot 6.02 \cdot 10^{23}}{0.0635} \approx 8.4 \cdot 10^{28} \, \text{m}^{-3}$$

Substituting the known constants ($e \approx 1.6 \cdot 10^{-19} \, \text{C}$, $m_e \approx 9.1 \cdot 10^{-31} \, \text{kg}$) into the formula for $\tau$:
$$\tau = \frac{5.8 \cdot 10^7 \cdot 9.1 \cdot 10^{-31}}{(1.6 \cdot 10^{-19})^2 \cdot 8.4 \cdot 10^{28}} = \frac{52.78 \cdot 10^{-24}}{2.56 \cdot 10^{-38} \cdot 8.4 \cdot 10^{28}} \approx 2.45 \cdot 10^{-14} \, \text{s}$$

$\textit{Note: A mathematical calculation based on modern reference data gives a value of $2.4 \cdot 10^{-14} \, \text{s}$. The answer provided in the textbook ($2.4 \cdot 10^{-15} \, \text{s}$) contains a typo by one order of magnitude.}$