Sulphur dioxide functions both as, a oxidising and as a reducing agent. In which one of the following reactions, it is an oxidising agent ?

An oxidant is that whose oxidation number decreases in a redox reaction by gain of an electron or electrons. Writing oxidation number of Sulphur atoms in the given answers, we have :

$$\begin{gathered} \left(1\right) {\mathrm{PbO}}_2+{\mathrm{SO}}_2⟶ {\mathrm{PbSO}}_4 [ \mathrm{S} :+4 \mathrm{to}+6] \mathrm{reducing} \mathrm{agent} \\ \begin{array}{l}\left(2\right) {\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2{\mathrm{SO}}_3 [ \mathrm{S} :\mathrm{No} \mathrm{change} \mathrm{in} \mathrm{o}.\mathrm{s} ]\\ \left(3\right) 3\mathrm{Fe}+{\mathrm{SO}}_2⟶2\mathrm{FeO}+\mathrm{FeS} [\mathrm{S} :+4 \mathrm{to} -2] \mathrm{oxidising} \mathrm{agent}\\ \left(4\right) 2\mathrm{NaOH}+{\mathrm{SO}}_2⟶{\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O} [\mathrm{S}:\mathrm{No} \mathrm{change} \mathrm{in} \mathrm{o}.\mathrm{s}]\end{array} \end{gathered}$$

From above, we see that the oxidation number of sulphur decreases from +4 to -2 in reaction (3). So, in this reaction, ${\text{SO}}_2$ acts as an oxidant.