=>

Since we have n = ab, we have nab = n^2 and 100 < n^2 < 200.
Perfect squares between 100 and 200 are 11^2 = 121, 12^2 = 144, 13^2 = 169 and 14^2 = 196.
n = 14 is the only one among those numbers, 11, 12, 13 and 14 which is a product of two different prime integers.
Thus we have a=2, b=7 or a=7, b=2.
Thus n - a - b = n - ( a + b ) = 14 - (2 + 7) = 14 - 9 = 5.

Therefore, B is the answer.
Answer: B

$$Given\ that\ n=ab,\ and\ nab=n^2\ and\ 100<n^2<200$$
Perfect squares between 100 and 200 are $$11^2=121,\ \ 12^2=144,\ \ 13^2=169\ and\ 14^2=196$$

Product of 2 numbers that gives n =>
For 11 = 1 * 11 and 1 is not a prime numbers
For n = 12 = 2 * 6 , 3 * 4 This is not a product of two different prime integers
For n = 13 = 1 * 13 and 1 is not a prime number
For n = 14 = 2 * 7 or 7 * 2 This is a product of two different prime integers

Therefore, we have a = 2; b = 7 or a = 7; b = 2
Hence, n - a - b = 14 - 2 - 7
= 12 - 7
= 5
Answer = option B