A Barrel is completely filled with acetone. 20% of the acetone is removed and replaced with water. This process is repeated two more times. What percentage of the resulting mixture is acetone?

Let the initial amount of acetone be 100 units.

After the first process:
Acetone remaining = 100 - (20% of 100) = 100 - 20 = 80 units
Water added = 20 units
Total mixture = 100 units

After the second process:
Acetone remaining = 80 - (20% of 80) = 80 - 16 = 64 units
Water added = 16 units
Total mixture = 80 units

After the third process:
Acetone remaining = 64 - (20% of 64) = 64 - 12.8 = 51.2 units
Water added = 12.8 units
Total mixture = 64 units

Percentage of acetone in the resulting mixture = (51.2 / 100) * 100 = 51.2%

Therefore, the correct option is A. 51.2