The whole difficulty lives in the word median. When a problem says "make one location as profitable as possible," the reflex is to make every other location as small as possible and pour the rest into that one. That reflex is right when you are maximizing the single largest value, but here you are maximizing the median, and the median is a rank, not a free magnitude. With seven locations it is the fourth profit when they are lined up in order, so three locations must rank above it. You cannot make those three small, because each of them has to be larger than the median itself.
Line up the seven distinct profits from smallest to largest. The median is the fourth one; call it M. To push M as high as possible, force both sides to their limits. The three locations below the median should be as small as distinct positive whole numbers allow, which is 1, 2, and 3, costing only 6 of the 56 million. The three locations above the median cannot be smaller than M and must be distinct, so the cheapest they can be is M + 1, M + 2, and M + 3. Everything is now pinned to M, so add it up: 1 + 2 + 3 + M + (M + 1) + (M + 2) + (M + 3) = 4M + 12, and that must equal 56. Solving gives 4M = 44, so M = 11.
Confirm that 11 is reachable and that nothing higher is. The set 1, 2, 3, 11, 12, 13, 14 uses seven different whole numbers, sums to 56, and has 11 as its middle value. Try to push the median to 12, and even the cheapest legal set, 1, 2, 3, 12, 13, 14, 15, already sums to 60, which overshoots 56. So 11 is the greatest possible median, choice (C).
The reusable method: to maximize a value defined by its rank, identify which other values that rank forces to be larger or smaller, push those to their tightest legal floor, push the freely shrinkable values to their floor too, set the total to the given sum, and solve. The discipline is to ask which constraint actually binds at the extreme, rather than reaching for the generic "minimize everything else" move, because that move maximizes the largest element instead of the ranked one.