$3 \times 2 = 6$
$4 \times 2 = 8$
I guess a question can't get easier than this- (D) choice. (Also, we can simply take the greatest value among choice for this question)
[There are $6$ resources and all of them must be in use for deadlock. If the system has no other resource dependence, $N=4$ cannot lead to a deadlock. But if $N=4$, the system can be in deadlock in presence of other dependencies.
Why $N=3$ cannot cause deadlock? It can cause deadlock, only if the system is already in deadlock and so the deadlock is independent of the considered resource. Till $N=3,$ all requests for considered resource will always be satisfied and hence there won't be a waiting and hence no deadlock with respect to the considered resource. ]