Answer is (A).
1. As soon as $N2-N3$ reduces to $2$,both $N2$ and $N3$ instantly updates their distance to $N3$ and $N2$ to $2$ respectively. So, $N2$: $(1, 0, 2, 7, 3)$, $N3$: $(7, 2, 0, 2, 6)$ becomes this.
After this starts first round of update in which each node shares its table with their respective neighbors ONLY. BUT KEEP IN MIND THAT ONLY OLD TABLES WILL BE SHARED.What I mean is tables that will be used for updation at this moment contain the values as $N1$$:$ $(0, 1, 7, 8, 4)$,$N2:$ $(1, 0, 2, 7, 3)$,$N3$: $(7, 2, 0, 2, 6)$,$N4:$ $(8, 7, 2, 0, 4)$,$N5:$ $(4, 3, 6, 4, 0)$.
SEE at this time all the entries are old EXCEPT in $N2$ and $N3$ where value changes to $2$ instead of $6$.
Question asks for $N3$. So focus on that.
N3 receives tables from $N2:$ $(1, 0, 2, 7, 3)$ and $N4:$ $(8, 7, 2, 0, 4)$. Using THIS ONLY original $N3:$ $(7, 2, 0, 2, 6)$ updates to $N3(3,2,0,2,5)$. (For updation and forming the tables for this refer FOROUZAN.)
So, answer is (A).