2021 Fall AMC 10A Problem 13

Note that for this restriction to hold, there must be $3$ balls of each color. There are $2^6 = 64$ ways to color the balls and $\binom{6}{3} = 20$ to choose which balls are white. The desired probability is therefore $\dfrac{20}{64} = \dfrac{5}{16}.$ Thus, D is the correct answer.