Since the first chip selected is black, Box B is rendered irrelevant: the first chip must be selected from A or C.
Between Boxes A and C, there are 3 black chips and 1 white chip:
BBBW
Once a black chip has been selected from A and C, the following chips remain:
BBW
Since there are now 2 black chips and 1 white chip -- for a total of 3 chips -- the probability that the next chip selected is black = 2/3.

I agree with your answer. Is it proven that the correct answer is D?

I've highlighted the error in your solution.

P(the selected chip is black) cannot equal 2/3. Here's why:
We have 1 all-white box, 1 all-black box, and 1 50-50 box.
Since white balls and black balls are equally represented, it must be the case that: P(we select a white ball) = P(we select a black ball)
Since P(we select a white ball) + P(we select a black ball) = 1, it must be the case that P(we select a white ball) = P(we select a black ball) = 1/2

Someone clarify my doubt below:
The case in question is the probability of selecting one black chip on first pick and second black chip on the second pick.
Since the question mentions 'same box', there is only 1 box where this is possible i.e., box A where both the chips are black.
Rest all other cases don't qualify towards the probability.
Therefore that evaluates as 1 out 3 boxes. Hence probability is 1/3. What's wrong with my logic?

it could pick the black form box C alike

got it, thanks!

The clue is in the question. Remember the AND/OR rule. Multiply probabilities when you see ‘and’. Add probabilities when you see ‘or’.

Think about the process. Draw a picture if it helps!

Let’s break it down. Ted has essentially 3 options.

Option 1 is that he chooses the first box (call it Box A) and the black chip inside.
Option 2 is that he chooses the second box (Box B) and the black chip inside.
Option 3 is that he chooses the third box (Box C) and the black chip inside.

Those options can be broken down using the AND/OR rule. Ted can choose either Box A, or Box B, or Box C.

So, the schema looks like:

(Box A * black chip) + (Box B * black chip) + (Box C * black chip)

The probability of choosing (Box A), (Box B) or (Box C) is (⅓) respectively.

The probability of choosing a black chip in box A is 100% (there is only one black chip in Box A).
The probability of choosing a black chip in box B is 0% (there are no black chips in Box B).
The probability of choosing a black chip in box C is 50% (half of the chips in Box C are black).

Therefore, plugging in the values into the formula gives you:

(⅓ * 1) + (⅓ * 0) + (⅓ * ½) = (⅓) + (⅙) =(½)

This is a relatively straightforward way to think about the question. If you are familiar with the AND/OR rule in probability, and how the rule applies, you can easily break this problem down and solve it within 60 seconds.

By calculating the eventualities iteratively, you could be stuck on this question for 90-120 seconds or even longer.

By the way… this is a layman's explanation of Bayes’ theorem and the probability formula. You will certainly save time on questions like this if you have a solid understanding of that formula and how to use it.

The Ultimate GRE Cheat Sheet has a comprehensive chapter on probability rules and time saving techniques. You must know these to get a top score on the GRE Quant. Go to Ultimate Tuition to download your copy.