Step 1: Understand the first reaction.
The starting compound is chlorobenzene.
The reagents
\[
HNO_3,\ \text{Conc. }H_2SO_4
\]
cause nitration of chlorobenzene.
Chlorine is deactivating but ortho-para directing due to resonance donation.
Therefore, nitration gives mainly ortho- and para-nitrochlorobenzene.
Due to steric hindrance at the ortho position, the para product is the major product.
So, the major nitration product is
\[
p\text{-nitrochlorobenzene}
\]
Step 2: Understand the effect of \(NaOH\) at \(443K\).
The presence of a nitro group at ortho or para position activates the aryl chloride toward nucleophilic substitution.
Thus, \(p\)-nitrochlorobenzene reacts with \(NaOH\) at \(443K\).
The chlorine atom is replaced by hydroxide ion.
Hence,
\[
p\text{-nitrochlorobenzene}
\xrightarrow{NaOH,\ 443K}
p\text{-nitrophenoxide}
\]
Step 3: Acidification step.
On treatment with
\[
H_3O^+
\]
the phenoxide ion is protonated to give phenol derivative.
Thus,
\[
p\text{-nitrophenoxide}
\xrightarrow{H_3O^+}
p\text{-nitrophenol}
\]
Step 4: Final conclusion.
Therefore, the major product is
\[
\boxed{4\text{-Nitrophenol}}
\]
Hence, the correct option is (4).