Let total marbles = x
Target question: how many white marbles are in the jar?
Therefore, black marbles = 2x/3 and white marbles = 1 - (2x/3) = x/3

Statement 1: If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.
Probability of ball drawn both to be black = 5/12
$$\left(\frac{\frac{2x}{3}}{x}\right)\cdot\left(\frac{\left(\frac{2x}{3}-1\right)}{x-1}\right)=\frac{5}{12}$$
$$\left(\frac{2x}{3}\cdot\frac{1}{x}\right)\cdot\left(\frac{2x-3}{3}\right)\cdot\left(\frac{1}{x-1}\right)=\frac{5}{12}$$
$$\left(\frac{2}{3}\right)\cdot\left(\frac{2x-3}{3x-3}\right)=\frac{5}{12}$$
$$\frac{4x-6}{9x-9}=\frac{5}{12}$$
$$12\left(4x-6\right)=5\left(9x-9\right)$$
$$48x-72=45x-45$$
$$48x-45x=-45+72$$
$$3x=27$$
$$x=\frac{27}{3}=9$$
From the question stem,
$$White=\frac{x}{3}=\frac{9}{3}=3$$
White = 3 marbles. Therefore, statement 1 is sufficient.

Statement 2: If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.
If (white marbles - 1) probability of next ball being white = 1/4
$$\frac{\left(\frac{x}{3}-1\right)}{x-1}=\frac{1}{4}$$
$$\frac{\left(x-3\right)}{3}\cdot\frac{1}{x-1}=\frac{1}{4}$$
$$\frac{x-3}{3x-3}=\frac{1}{4}$$
$$4\left(x-3\right)=3x-3$$
$$4x-12=3x-3$$
$$4x-3x=-3+12$$
$$x=9$$
From the question stem,
$$White=\frac{x}{3}=\frac{9}{3}=3$$
White = 3 marbles. Therefore, statement 2 is sufficient.

Conclusively, since each statement alone is SUFFICIENT, answer = option D.