Step 1: Volume stays fixed:
Stretching only reshapes the wire. Its volume does not change. So tripling the length cuts the area to one third: $A' = A/3$.
Step 2: Resistance in terms of length:
Write the resistance as $R = \rho \dfrac{l}{A} = \rho \dfrac{l^2}{\text{Volume}}$.
With a fixed volume, $R$ is proportional to $l^2$. The length becomes $3l$, so the resistance becomes $3^2 = 9$ times larger.
\[ R' = 9R \]
Step 3: Current after stretching:
The same voltage $V$ now drives a resistance of $9R$. So the current drops to one ninth.
\[ I' = \frac{V}{9R} = \frac{I}{9} \]
Step 4: Drift speed from current:
Use $I = n e A v_d$, so $v_d = \dfrac{I}{neA}$.
\[ v_d' = \frac{I/9}{n e (A/3)} = \frac{1}{9} \times 3 \times \frac{I}{neA} = \frac{v_d}{3} \]
This matches the field argument: $E' = V/3l = E/3$, so the drift speed falls to one third.
Final Answer:
The drift speed decreases to one third and the resistance increases to nine times.
\[ \boxed{v_d' = \frac{v_d}{3}, \quad R' = 9R} \]