1. Find the oxidation state change required for manganese

In permanganate, manganese is in the oxidation state: $\text{Mn in } \mathrm{MnO_4^-} = +7$

The question says the change in oxidation state of manganese occurs by 3. So manganese must go from: $+7 \to +4$

This corresponds to formation of manganese dioxide: $\mathrm{MnO_4^- \to MnO_2}$

2. Recall reduction products of permanganate in different media

Permanganate gives different products depending on the medium:

• Acidic medium: $\mathrm{MnO_4^- \to Mn^{2+}}$ Change in oxidation state: $+7 \to +2 \quad (\Delta = 5)$

• Neutral or faintly alkaline medium: $\mathrm{MnO_4^- \to MnO_2}$ Change in oxidation state: $+7 \to +4 \quad (\Delta = 3)$

• Strongly alkaline medium: $\mathrm{MnO_4^- \to MnO_4^{2-}}$ Change in oxidation state: $+7 \to +6 \quad (\Delta = 1)$

3. Match with the condition in the question

Since the oxidation state change is 3, the reaction must occur in the medium where permanganate is reduced to $\mathrm{MnO_2}$.

That happens in: $\text{neutral or faintly alkaline medium}$

4. Check the options

• A: aqueous neutral — correct, since neutral medium gives $\mathrm{MnO_2}$.
• B: both aqueous acidic and faintly alkaline — incorrect, acidic gives change of 5.
• C: both aqueous acidic and neutral — incorrect, acidic again gives change of 5.
• D: aqueous acidic — incorrect.

5. Final answer

The medium that favours oxidation-state change of manganese by 3 is aqueous neutral.

So the correct option is: $\boxed{\text{A}}$