To solve this problem, we'll break it down into steps and consider the different scenarios that could happen, based on the selection of balls from Box A to Box B and then the selection from Box B.

**Step 1: Selection from Box A to Box B**
Initially, Box A contains 6 white and 4 black balls, and Box B contains 2 white and 2 black balls.

When two balls are selected from Box A and placed into Box B, there are three possible outcomes:

**Two white balls are selected from Box A:** Probability = 6/10 * 5/9
**One white and one black ball are selected from Box A:** Probability = 6/10 * 4/9 + 4/10 * 6/9
**Two black balls are selected from Box A:** Probability = 4/10 * 3/9

**Step 2: Selection from Box B**
After the transfer, depending on which balls were added, we will have different compositions of Box B and thus different probabilities for selecting exactly one white ball out of two selected balls.

**If two white balls were added to Box B:** Box B will have 4 white and 2 black balls. The probability of selecting exactly one white ball is given by selecting one white out of four and one black out of two.
**If one white and one black ball were added to Box B:** Box B will have 3 white and 3 black balls. The probability of selecting exactly one white ball is given by selecting one white out of three and one black out of three, or the reverse.
**If two black balls were added to Box B:** Box B will have 2 white and 4 black balls. The probability of selecting exactly one white ball is given by selecting one white out of two and one black out of four.

∴ The probability that exactly one of the two balls selected from Box B is white is approximately 56.9%