The correct answer is $\boxed{\text{B) 0.18}}$.
The probability of a student failing in both the papers is the probability of failing in Paper 1 and failing in Paper 2, given that the student has failed in Paper 2. This can be calculated as follows:
$$P(\text{fail in both}) = P(\text{fail in 1 and 2}) = P(\text{fail in 1}) \cdot P(\text{fail in 2} | \text{fail in 1})$$
We are given that $P(\text{fail in 1}) = 0.3$ and $P(\text{fail in 2} | \text{fail in 1}) = 0.6$. Substituting these values into the equation, we get:
$$P(\text{fail in both}) = 0.3 \cdot 0.6 = 0.18$$
Option A is incorrect because it is the probability of failing in Paper 1, not the probability of failing in both papers. Option C is incorrect because it is the probability of failing in Paper 2, not the probability of failing in both papers. Option D is incorrect because it is the probability of failing in Paper 1, given that the student has failed in Paper 2, not the probability of failing in both papers.