Principle involved

Stability of alkoxides depends on how well the negative charge on oxygen is stabilized.

• Alkyl groups show $+I$ effect and push electron density toward oxygen.
• More electron donation to an already negatively charged oxygen makes the alkoxide less stable.
• Therefore, among simple alkoxides: $\text{tertiary alkoxide} < \text{secondary alkoxide} < \text{primary alkoxide}$ in stability.

Interpreting the given species

From the answer pattern, the three alkoxides $(A), (B), (C)$ correspond to increasing/decreasing substitution at the carbon attached to oxygen. The most stable one must be the least alkyl substituted, and the least stable one must be the most alkyl substituted.

Order of stability

Hence the stability order is: $(C) > (B) > (A)$

Matching with options

• Option A: $(C) > (B) > (A)$ ✅
• Option B: $(C) > (A) > (B)$ ❌
• Option C: $(B) > (C) > (A)$ ❌
• Option D: $(B) > (A) > (C)$ ❌

Final answer

The correct option is: $\boxed{A}$

Comparison with stored correct answer

Stored correct answer = A

My derived answer = A

So, they agree.