The above solutions are strong, but I can highlight my train of thought behind it. The solution isn't optimal, but the thinking might be helpful.
40% of all marbles are black. We will pick three marbles one after the other (no replacement means, we put back the marble each time), and determine the probability of getting one of the three as black, two of the three as black, or all three as black.
All three is easy. 40 / 100 = 0.4; so, 0.4*0.4*0.4 = 0.064. (1)
Two out of three black will mean 0.4*0.4*0.6 (as 60% is the probability of the marble not being black). That's 0.096. But that white marble can be any of the three - the first, the second or the third. So, 0.096*3 = 0.288. (2)
One out of three black will mean 0.4*0.6*0.6 = 0.144. The black marble can be any of the three, so, 0.144*3 = 0.432 (3)
Together, (1) + (2) + (3) = 0.064 + 0.288 + 0.432 = 0.784, or E. Perhaps using fractions like 2/5 can make the calculation easier.