for(i = 1; i< n ; i++) //O(n)
for(j = 1; j<i; j++) //O(i) = O(n)
if(i < j == 0) // Means "i < j is False?" which is true for every instance. So, proceed.
for(k = 0; k<j; k++) // O(j) = O(n)
sum++;
Hence, $O(n^3)$
The if-condition will never be false, so the best case would be $\Omega (n^3)$. Hence, $\Theta (n^3)$
Option D