File size = 2100 when it goes to network layer 20 bytes are added so it becomes 2120B but this is just for understanding as we can solve it without considering IP overhead because fragmentation is done on data not headers.
At sender S : 2100 B to intermediate router P(1200 bytes without header) to intermediate router Q(400 B w/o header) to destination D.
S to P :
fragement 1 : 1200 fragment 2 : 900 to P
P to Q :
for 1200B fragment
F1 : 400(data length in bytes) offset : 0
F2 : 400 offset : 400/8=50
F3 : 400 offset : 100
for 900B fragment
F4: 400 offset : 150
F5: 400 offset : 200
F6: 100 offset : 250 last fragment
In other way 250*8 = 2000 bytes ahead of this last fragment (from 0 to 1999) and sequence number of last fragment is 2000.
So 250 shoud be answer.