Set S = {2, 3, 8, 11, x, y}
$$Mean\ of\ S=\frac{2+3+8+11+x+y}{6}$$
$$7=\frac{24+x+y}{6}$$
Median = 5.5 and x<y
From the mean of S
$$24+x+y=42$$
$$x+y=18$$
From the median, an average of 2 numbers in the middle = 5.5
$$i.e\ \frac{a+b}{2}=5.5\ and\ a+b=11$$
From the numbers in the set 3+8=11. This means that x and y might not be in the middle and since x<y, $$The\max imum\ possible\ value\ of\ x\ is\le3$$

Answer = option D