Step 1: What we have to show:
We need a model of motion of electrons in a wire, the formula for the current, a reason for the instant start of current, and a ratio of drift speeds in two wires.
Step 2: Part (i), steady average velocity:
Think of an electron in the wire before the field is switched on. It moves randomly, so its average velocity is zero.
When the field $E$ is applied, each electron feels an extra force $eE$ opposite to the field. Between collisions, its velocity changes by $eE t/m$.
Collisions with ions destroy this gain. If the mean time between collisions is $\tau$, then the average extra velocity is the velocity gained in the mean time $\tau$:
\[ v_d = \dfrac{eE\tau}{m}\ \text{(magnitude)} \]
This value does not depend on time. The field keeps adding velocity and the collisions keep removing it, so a steady balance is reached.
Step 3: Current from drift velocity (via current density):
Take $n$ free electrons per unit volume. In unit time, the electrons that cross unit area are those in a prism of length $v_d$. This gives the current density $j = n e v_d$.
The total current is $j$ times the area, so:
\[ I = j A = n e A v_d \]
Step 4: Part (ii), the instant current:
The electrons already sit all along the wire, like water already filling a pipe. Closing the switch sets up the electric field along the whole wire at about $3\times10^8$ m/s.
All electrons start to drift together. The slow drift speed of a few mm per second only tells how far one electron moves, not how fast the signal travels.
Step 5: Two copper wires in series:
Same current in both wires. Same material means the same $n$. So $n e A_1 v_1 = n e A_2 v_2$, which gives $v_1/v_2 = A_2/A_1$.
With radii $3k$ and $2k$, the areas are in the ratio $9 : 4$, so $v_1/v_2 = 4/9$.
Final Answer:
$I = neAv_d$ with $v_d = eE\tau/m$ in magnitude. Current appears at once since the field spreads at nearly light speed. The drift velocities are in the ratio
\[ \boxed{4 : 9} \]