First, as soon as $N1-N2$ goes down, $N2$ and $N1$ both update that entry in their tables as infinity. So $N2$ at this moment will be $N2(\text{inf},0,2,\_,\_).$ I have left blank because that details are not important.
Now for $N3$ to get updated in the subsequent round it will get tables from $N2$ and $N4$ only. But first we need to find the $N4$ calculated in previous update. So in previous question $N4$ received updates from $N3$ and $N5$ which are $N3: (7, 6, 0, 2, 6),N5: (4, 3, 6, 4, 0).$
NOW THIS IS VERY IMPORTANT AS WHY N4 DID NOT GET UPDATED TABLES FROM N3. SO ANSWER IS THAT these tables were shared at the same moment and so in a particular round of update old values of all the tables are used and not the updated values.
$N3$ was updates AFTER IT PASSED ITS OLD table to its neighbors AS WHY WOULD $N4$ WAIT FOR $N3$ to GET UPDATED first !!! So N4 will update its table (in prev question) to N4(8,7,2,0,4).
See here path to $N1$ exists via $N5$ and not via $N3$ because when table was shared by $N3$ it contained path to $N1$ as $7$ and $N1$ via $N3$ sums to $7+2 =9.$ Now when $N3$ receives tables from $N2(\text{inf},0,\_,\_,\_)$ and $N4(8,7,2,0,4).$
At first it will see its distance to $N1$ as "Inf" and NOT $3$ because "inf" is the new distance with the same Next hop $N2$ (If next hop is same, new entry is updated even though it is larger than previous entry for the same NEXT HOP).
But at the same time it sees distance to $N1$ from $N4$ as $8$ and so updates with the value $(N3-N4 + N4-N1)= (2+8)=10.$ So $N3-N1$ distance in $N3(10,\_,0,\_,\_)$ is $10.$
So, answer is (C)
Reference: http://www.cs.princeton.edu/courses/archive/spr11/cos461/docs/lec14-distvector.pdf