1. We need to identify the product A formed when thiosulphate ion, $\mathrm{S_2O_3^{2-}}$, is oxidized by permanganate, $\mathrm{MnO_4^-}$, in mildly alkaline medium.

2. Thiosulphate commonly undergoes oxidation to tetrathionate under mild oxidizing conditions: $2\,\mathrm{S_2O_3^{2-}} \rightarrow \mathrm{S_4O_6^{2-}} + 2e^-$ So here, $A = \mathrm{S_4O_6^{2-}} \quad \text{(tetrathionate ion)}$

3. Now find the oxidation states of sulphur in $\mathrm{S_4O_6^{2-}}$.

   The structure of tetrathionate can be viewed as: $\mathrm{^-O_3S - S - S - SO_3^-}$ There are two kinds of sulphur atoms:

• the two terminal sulphur atoms bonded to oxygen are each at oxidation state $+5$
• the two inner sulphur atoms are each at oxidation state $0$

4. Check by average oxidation state method: Let average oxidation state of sulphur be $x$. Then, $4x + 6(-2) = -2$ $4x - 12 = -2$ $4x = 10$ $x = \frac{5}{2}$ This is only the average, not the oxidation state of each sulphur atom.

5. Since the question asks for the oxidation state of sulphur in product $A$, and in tetrathionate sulphur exists in two oxidation states, the oxidized sulphur atoms are at $+5$.

Thus, the required oxidation state is: $\boxed{5}$