On second thought, the answer is E, not C as I originally proposed. Sorry.
Let's assume X1 = 8, then X2 = 12. In this case, the median of the series is 12 for sure if we combine both conditions. In this case there are three possibilities for X2 in the series of X2, X3, and X4: biggest, smallest, or the middle. If X2 is the smallest, then either X3 or X4 is 12. So the median of the 4 numbers is 12. However, is X2 is the middle number, then we do not know the median value of the 4 numbers because we have to take an average of X2=12 with another undecided number. It could be X1=10 or X3=11. In sufficient.