This is $N$ added itself $N$ times. So it is $N^2$ . Even if you consider as sum of $O(1) + O(2) + \ldots +O(n-1) + O(N)$. it will add up to $N^2$
So, the answer is:
$(A)\; O(N)$ this is false.
$(B, C, D, E )$ All of this is true. We have $N^2$ here, so all options apart from $(A)$ are correct.
In fact $B = D = E$ this three options are same. and $N^3$ is always the upper bound of $N^{2}$. So $O(N^3)$ is also true.
PS: $\displaystyle{} \sum (K=1 \;\text{to}\; n)O(K)$ is never equal to $n$ functions. It is always equal to one single function.
It can be written as $c.1+c.2+c.3$ and so on which results in $O(N^2)$ (source Cormen)