Step 1: Identify the species in the reaction.
The sulphur containing reactant is the sulphite ion $SO_3^{2-}$, and it reacts with bromine $Br_2$ in the presence of water. We must find the oxidation state of sulphur in the sulphur containing product.
Step 2: Recognise bromine as an oxidant.
Bromine is a good oxidizing agent. In water it pulls electrons from sulphite, oxidizing it. So sulphur in the product will be in a higher oxidation state than in $SO_3^{2-}$.
Step 3: Find the oxidation state of sulphur in sulphite.
In $SO_3^{2-}$, let sulphur be $x$. \[ x + 3(-2) = -2 \] \[ x - 6 = -2 \] \[ x = +4 \]
Step 4: Identify the oxidized product.
Oxidation of sulphite by bromine in water gives the sulphate ion $SO_4^{2-}$, which is the stable higher state. \[ SO_3^{2-} + Br_2 + H_2O \longrightarrow SO_4^{2-} + 2Br^- + 2H^+ \]
Step 5: Find the oxidation state of sulphur in sulphate.
In $SO_4^{2-}$, let sulphur be $y$. \[ y + 4(-2) = -2 \] \[ y - 8 = -2 \] \[ y = +6 \]
Step 6: State the answer.
The oxidation state of sulphur in the product $SO_4^{2-}$ is $+6$, matching the key.
\[ \boxed{+6} \]