To solve this, let’s break down the problem:

1. Initial Mixture:

   • The milkman starts with 120 liters of milk and 30 liters of water.
   • Total volume of the mixture = $120 + 30 = 150$ liters.

2. Mixture Sold:

   • He sells one-third of the total mixture, which is $\frac{1}{3} \times 150 = 50$ liters.
   • The 50 liters sold will contain both milk and water in the same ratio as the original mixture.

   The initial milk-to-water ratio is: $\frac{120}{30} = 4:1$

3. Amount of Milk and Water in 50 Liters Sold:

   • Water in the 50 liters = $\frac{1}{5} \times 50 = 10$ liters
   • Milk in the 50 liters = $\frac{4}{5} \times 50 = 40$ liters

4. Remaining Mixture:

   • After selling 50 liters, he has:
     • Water: $30 - 10 = 20$ liters
     • Milk: $120 - 40 = 80$ liters
   • Remaining mixture = $80 + 20 = 100$ liters.

5. Adding Water to Compensate:

   • He adds water to bring the mixture back to the original total volume of 150 liters.
   • Required amount of water to add = $150 - 100 = 50$ liters.

6. Final Mixture Composition:

   • Total water = $20 + 50 = 70$ liters
   • Total milk = 80 liters

7. Final Ratio of Water to Milk: $\text{Water : Milk} = 70 : 80 = 7 : 8$

Thus, the correct answer is not in the options given.

If any questions or details needed?
Here are 5 questions to expand understanding:

1. How to calculate the proportion of substances when the ratio is given?
2. Why did the milk and water ratio remain the same in each fraction of the mixture sold?
3. What other ways could the milkman adjust the mixture to compensate for the water?
4. How can we determine the final ratios when scaling quantities up or down?
5. In what real-life situations are these mixture ratio calculations applicable?

Tip: Always verify the volume of individual components before and after operations in mixture problems for accuracy.