Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is

Bag I contains 3 red, 4 black, and 3 white balls (total 10). Bag II contains 2 red, 5 black, and 2 white balls (total 9). One ball is transferred from Bag I to Bag II, then a ball is drawn from Bag II and found to be black. We need the probability that the transferred ball was red.

We use Bayes' theorem. Let $$R, B, W$$ denote the events that a red, black, or white ball was transferred, respectively.

The prior probabilities are $$P(R) = \frac{3}{10}$$, $$P(B) = \frac{4}{10}$$, $$P(W) = \frac{3}{10}$$.

After transfer, Bag II has 10 balls. The probability of drawing a black ball from Bag II given each case:
Case 1: If red was transferred, Bag II has 5 black out of 10, so $$P(\text{black}|R) = \frac{5}{10} = \frac{1}{2}$$.
Case 2: If black was transferred, Bag II has 6 black out of 10, so $$P(\text{black}|B) = \frac{6}{10} = \frac{3}{5}$$.
Case 3: If white was transferred, Bag II has 5 black out of 10, so $$P(\text{black}|W) = \frac{5}{10} = \frac{1}{2}$$.

By the law of total probability: $$P(\text{black}) = \frac{3}{10} \cdot \frac{1}{2} + \frac{4}{10} \cdot \frac{3}{5} + \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20} + \frac{12}{50} + \frac{3}{20} = \frac{3}{20} + \frac{6}{25} + \frac{3}{20}$$.

Converting to a common denominator of 100: $$= \frac{15}{100} + \frac{24}{100} + \frac{15}{100} = \frac{54}{100} = \frac{27}{50}$$.

By Bayes' theorem: $$P(R|\text{black}) = \frac{P(\text{black}|R) \cdot P(R)}{P(\text{black})} = \frac{\frac{1}{2} \cdot \frac{3}{10}}{\frac{27}{50}} = \frac{\frac{3}{20}}{\frac{27}{50}} = \frac{3}{20} \cdot \frac{50}{27} = \frac{150}{540} = \frac{5}{18}$$.

Hence, the correct answer is Option B.