The correct answer is $\boxed{\text{C. 30%}}$.
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
In this case, event A is picking a red ball, and event B is mixing the contents of Bags A and B. The probability of picking a red ball from Bag A is $0.2$, and the probability of picking a red ball from Bag B is $0.3$. The probability of mixing the contents of Bags A and B is $1$, since it is certain that the contents will be mixed. Therefore, the conditional probability of picking a red ball given that the contents of Bags A and B have been mixed is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2 + 0.3}{1} = 0.3$$
Therefore, the probability of picking a red ball is $\boxed{\text{30%}}$.
Option A is incorrect because it is the probability of picking a red ball from Bag A, not from the mixture of Bags A and B. Option B is incorrect because it is the average of the probabilities of picking a red ball from Bags A and B, but these probabilities are not equally likely. Option D is incorrect because it is the probability of picking a red ball from Bag B, not from the mixture of Bags A and B.