Br${}_2$/AlBr${}_3$ on $p$-nitroethylbenzene: the ethyl group is $o,p$-directing while $-$NO${}_2$ is $m$-directing. The $-$NO${}_2$ blocks the para position, so bromination occurs ortho to the ethyl group (and meta to NO${}_2$), giving compound A.

Sn/HCl reduces $-$NO${}_2$ to $-$NH${}_2$ (compound B); NaNO${}_2$/HCl at $0$-$5^{\circ }$C diazotises the amine (compound C); ethanol reduces the diazonium salt to replace the $-$N${}_2^{+}$ group by $-$H (compound D), now an ethyl/bromo-disubstituted benzene with the two substituents ortho to each other.

Alkaline KMnO${}_4$ followed by acidification oxidises the ethyl group entirely to $-$COOH; the bromine on the ring is retained. The product E is $o$-bromobenzoic acid (the $-$Br and $-$COOH are on adjacent ring carbons).

Hence option (2).