1. Initially, container X has 100 ml of milk and container Y has 100 ml of water.

2. Step 1: Transfer 20 ml of milk from X to Y. After this transfer, container X has 80 ml of milk. The amount of liquid in container Y becomes 120 ml (100 ml of water + 20 ml of milk). The ratio of milk in container Y is: Milk in Y = $20 \div 120 = \frac{1}{6}$ Thus, the ratio of water in Y is: Water in Y = $100 \div 120 = \frac{5}{6}$

3. Step 2: Transfer 20 ml of the mixture from Y back to X. The mixture in Y consists of 1/6 milk and 5/6 water. So, in the 20 ml transferred: Amount of milk = $(\frac{1}{6}) \times 20 = 3.33$ ml Amount of water = $(\frac{5}{6}) \times 20 = 16.67$ ml After transferring 3.33 ml of milk back to X, the milk in X becomes: 80 ml + 3.33 ml = 83.33 ml The amount of water in container Y after the transfer becomes: 100 ml - 16.67 ml = 83.33 ml

4. Proportions: The proportion of milk in X is: $m = 83.33 \div 100 = 0.8333$ The proportion of water in Y is: $n = 83.33 \div 100 = 0.8333$

Since $m = n$, the correct answer is $m = n$.