The inner most loop will run for O(log(m)) as m is getting halved in every iteration of the while loop. Now, the loop with iterator j runs for n/1000 which is independent of the value of i. So, j runs for O(n) every time and then the outermost loop with iterator as i runs for O(n) time.
Therefore, total complexity = $O(n^{2}log(m))$ . So, (a) should be the correct answer.
