There are 5 essentially different ways to distribute the black balls. In each case I'll count the essentially different ways of distributing the white balls.
4, 0, 0, 0
The possible distinguishable ways of distributing the white balls are: 2-0-0-0, 1-1-0-0, 0-2-0-0 and 0-1-1-0. So 4 is the number.
3, 1, 0, 0
Here we can do it like this: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-2-0-0, 0-1-1-0, 0-0-2-0, 0-0-1-1. So 7.
2, 2, 0, 0
Once again: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-0-2-0, 0-0-1-1. 5 ways.
2, 1, 1, 0
And again: 2-0-0-0, 1-1-0-0, 1-0-0-1, 0-2-0-0, 0-1-1-0, 0-1-0-1, 0-0-0-2. 7 ways.
1, 1, 1, 1
Lastly: 2-0-0-0, 1-1-0-0. 2 ways.
In total, $4+7+5+7+2 = 25$ ways to distribute the balls.
Case 1
You picked a white ball from the first box.
This happens with a chance of $3/5$. Putting it into the second box, now there are $5$ white and $4$ black balls. Now the chance to pick a white ball is $5/9$. So the total chance in case 1 is $3/5 \cdot 5/9 = 1/3$.
Case 2
You picked a black ball from the first box.
This happens with a chance of $2/5$. Putting it into the second box, now there are $4$ white and $5$ black balls. Now the chance to pick a white ball is $4/9$. So The total chance in case 2 is $2/5 \cdot 4/9 = 8/45$.
Combining the two cases, the chance is $$1/3 + 8/45 = 23/45 \approx 51\%.$$
Best Answer
There are $20$ boxes and $20$ balls in total, so each box must receive exactly one ball. The question is reduced to choosing which $10$ boxes get black balls and which $10$ get white balls, which is equivalent to choosing $10$ boxes to put the black balls in and filling the rest with white balls. The number of ways to choose $10$ boxes out of $20$ is $${20 \choose 10} = \frac{20!}{10!\cdot 10!},$$ which is the answer.