Given: Total volume of milk sample = $x$ liters Water in the sample = $50$% of $x$ = $x/2$ liters Milk in the sample = $x/2$ liters

Step 1: $1/3$rd of the sample is taken. Milk taken = $x/3$ liters Water in the taken portion = $50$% of $x/3$ = $x/6$ liters Milk in the taken portion = $x/6$ liters

Step 2: Equal amount of pure milk ($x/3$ liters) is added. Water in pure milk = $0$ liters Milk in pure milk = $x/3$ liters

Step 3: New mixture: Total volume = $(x/3) + (x/3) = 2x/3$ liters Total water = $(x/6) + 0 = x/6$ liters Total milk = $(x/6) + (x/3) = x/2$ liters

Step 4: Percentage of water in the new mixture: Percentage of water = (Total water / Total volume) $\times 100$ Percentage of water = $(x/6) / (2x/3) \times 100$ Percentage of water = $(x/6) \times (3/2x) \times 100 = 25$%

Answer: A. $25$%