Solution:

The answer is (D). The multiples of 3 between 1 and 20 are 6, so there are 14 numbers that are not multiples of 3. If we were to draw exactly those 14 numbers, their product would not be a multiple of 3, and even less of 12, so the minimum number $n$ of draws needed to ensure that the product is a multiple of 12 is greater than 14. If instead we draw 15 numbers, we will certainly have at least one multiple of 3. Since the even numbers between 1 and 20 are 10, and the odd ones are 10, with 15 draws we are guaranteed at least 5 even numbers. Thus the product will be a multiple of 3 and a multiple of $2^{5}=32$. In particular, it will be a multiple of 3 and of 4, and hence of 12. It follows that $n$ is exactly 15.