1. Identify the reacting species

In acidic medium, potassium permanganate oxidizes ferrous ion to ferric ion. Ferrous ammonium sulphate hexahydrate is Mohr’s salt: $\mathrm{FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O}$ Each formula unit provides one $\mathrm{Fe^{2+}}$.

2. Write the ionic redox equation in acidic medium

The standard reaction is: $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$

For 2 molecules of $\mathrm{KMnO_4}$, multiply by 2: $\mathrm{2MnO_4^- + 10Fe^{2+} + 16H^+ \rightarrow 2Mn^{2+} + 10Fe^{3+} + 8H_2O}$

So, from the redox process itself, 8 molecules of water are formed.

3. Account for water of crystallization from Mohr’s salt

To react with $2\,\mathrm{KMnO_4}$, we need $10$ $\mathrm{Fe^{2+}}$ ions, hence $10$ molecules of Mohr’s salt.

Each molecule of Mohr’s salt contains $6$ water molecules of crystallization. Thus total water contributed is: $10 \times 6 = 60$

4. Total number of water molecules produced/present in products side

Including the water formed in the redox reaction and the water released from the hydrated salt: $60 + 8 = 68$

5. Final answer

The number of water molecules for $2$ molecules of $\mathrm{KMnO_4}$ is: $\boxed{68}$