Big O notation is used to give the maximum upper bound for running time.This means that the algorithm will not take time greater than the binding function. Let's analyze each of these
A) (15^10) * n + 12099 In this polynomial we can safely ignore the constant part 12099, this leaves us with (15^10)*n. Now here too we can ignore (15^10) because it also a constant. This gives us a bound of Theta(n).This Theta (n) will always be bound by O(n^2).
B) n^1.8 will also always be bound by O(n^2).
C) n^3/(sqrt(n))= n^3/ n^0.5= n^2.5, You can clearly see that n^2.5 cannot be bound by n^2. This will increase much faster than n^2. Hence this cannot be O(n^2).
D) (2^20)*n, Similar to the first case this is linear in input as well. Hence it will also be bound by O(n^2).
I hope you've understood the approach.