A. The drift velocity of electrons decreases with the increase in...

Drift velocity $= \left(\dfrac{eτ}{m}\right)E$

${v}_{d} = \left(\dfrac{eτ}{m}\right)\left(\dfrac{\DeltaV}{l}\right)$

$\DeltaV =$ Potential difference applied across the wire As temperature increases, relaxation time decreases, hence ${V}_{d}$ decreases.
As per formula, ${V}_{d} ∝ \dfrac{1}{l}$

${v}_{d} = \dfrac{I}{neA}$, as it is not mentioned that current is at steady state neither it is mentioned that $n$ is constant for given conductor. So it can't be said that ${v}_{d}$ is inversely proportional to $A$.

$I = neA{v}_{d} = \dfrac{V}{R} = \dfrac{V}{ρl}A$

${v}_{d} = \dfrac{eEτ}{m}$

$τ$ decrease with temperature increase.

First and fourth statements are correct.

${v}_{d} = \dfrac{V}{ρlne}\left(E = \dfrac{V}{l}\right)$