Sulphur dioxide functions both as an 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 :

(a)      ${\mathrm{PbO}}_2+{\mathrm{SO}}_2⟶{\mathrm{PbSO}}_4 : \mathrm{change} \mathrm{in} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{S} :+4 \mathrm{to} +6$

(b)      ${\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2{\mathrm{SO}}_3 : \mathrm{NO} \mathrm{change} \mathrm{in} \mathrm{o}.\mathrm{s} \mathrm{of} \mathrm{S}$

(c)          $3\mathrm{Fe}+{\mathrm{SO}}_2⟶2\mathrm{FeO}+{\mathrm{FeS}}^{ } : \mathrm{change} \mathrm{in} \mathrm{o}.\mathrm{s} \mathrm{of} \mathrm{S} :+4 \mathrm{to} -2$

(d)      $2\mathrm{NaOH}+{\mathrm{SO}}_2⟶{\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O} : \mathrm{NO} \mathrm{change} \mathrm{in} \mathrm{o}.\mathrm{s} \mathrm{of} \mathrm{S}$

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