First calculate the probability that the larger number is n (1 <= n <= 6). This is exactly (2n-1)/36. Why? There are three possibilities:
The first die is exactly n, the second die is smaller -- So second die can be 1 ... n-1.
The second die is exactly n, the first die is smaller -- So the first die can be 1 ... n-1.
Both die are the same. This happens in only one way.
So total number of ways in which the larger number is n = 2n-1. Hence the probability that the larger number is n is (2n - 1)/36.
Consequently, the expected value of the larger of the two numbers is
Sum [n = 1 to 6] (2n -1)/36 * n
= 1/36 (Sum [n = 1 to 6] 2n^2 - Sum [n = 1 to 6] n)
You can compute this either directly or by using formulas for sequences. The answer is 161/36.
Since the expected loss is 50 cents, we know it is more likely to lose than to win. It is also impossible to get -50, and the total difference between the 2 possibilities is 2, 50/200=1/4, so the chance of winning is 1/4 and the chance of losing is 3/4. Since (1/4)*3=3/4, the amount of black balls are 3 times the amount of white balls, so k=15.
Hope this helps!! :)