Answer will be (C)
Tcp segment size = 900 +20 is 920 bytes
when it will go to IP layer at A, A's network layer will receive 920 as whole IP data and it will add 20 bytes of header
so IP packet size = 920 +20
=940 bytes
now H1-----R1
MTU=1024 bytes
so no need of fragmentation and (940 + 14) bytes are transferred
for R1--------R2
WHEN 940 comes it should be fragmented
data part is 920 so it is fragmented into 480 and 440 bytes
now (480+20+8)and (440+20+8) is sent
for R2-------B
again 920 is fragmented into 480 and 440
so,
(480+20+12)and (440 +20+12) are sent
hencelast fragment is 440+20+12 in which data is (440+20) =460 bytes
offset(how much data is ahead of the fragment) = 480/8 =60