$v _{ d }=\left(\frac{ e \tau}{ m }\right)\left(\frac{\Delta V }{\ell}\right)$

$\Delta V=$ Potential difference applied across the wire

As temperature increases, relaxation time decreases, hence $V _{ d }$ decreases.

As per formula, $V _{ d } \propto \frac{1}{\ell}$

$v _{ d }=\frac{ I }{\text { neA }}$, 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 $A$.

$I=n e A v_{d}=\frac{V}{R}=\frac{V}{\rho \ell} A$

$v _{ d }=\frac{ V }{\rho \ell \text { ne }} \quad\left( E =\frac{ V }{\ell}\right)$

$v _{ d }=\frac{ eE \tau}{ m }$

$\tau$ decrease with temperature increase.

First and fourth statements are correct.