The correct answer is $\boxed{\frac{3}{7}}$.
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 the event that the second ball removed is red, and event B is the event that the first ball removed is white. We are given that event B has already happened, so we need to calculate $P(A \cap B)$ and $P(B)$.
$P(A \cap B)$ is the probability that both the first ball removed is white and the second ball removed is red. This can be calculated as follows:
$$P(A \cap B) = \frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$$
$P(B)$ is the probability that the first ball removed is white. This can be calculated as follows:
$$P(B) = \frac{4}{7}$$
Therefore, the conditional probability that the second ball removed is red, given that the first ball removed is white, is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{7}}{\frac{4}{7}} = \frac{3}{7}$$