The crossover point where it makes sense to use the O(n) algorithm will occur when:
200 * n = n lg n
which occurs for n > 2^200. Most of us are not dealing with problems larger than can be represented in our universe, so cache efficiency matters, kids.
Note that this changes for other comparisons. Going from n maximally cache inefficient vs n^2 cache efficient is only worth it for n up to 200, and n lg n --> n^2 up to n about 2223.
Of course, in practice your algorithm won't be missing the cache on every access, so reality is somewhere between these values.