Concept:IP Datagram: Datagram contains data and header both.Offset: How many bytes ahead of this particular fragment.Each IP packet will have 1480Byte data and 20Bytes for headerCalculation:Total data we have to send is 4404Bytes - 20Bytes = 4384BytesNo of fragments \(= \frac{{4384}}{{1480}} = 3\)First fragment:148020Offset of first fragment: 0More fragment bit is 1 because there are two more fragments present.Second fragment:148020 Total 1480 Bytes are ahead of this fragment so offset \({\rm{}} = \frac{{1480}}{8} = 185\)More fragment bit is 1 because one more fragment is present.Third fragment:142420 Total 1480 + 1480 Bytes = 2960 Bytes are ahead of this fragment so offset \({\rm{}} = \frac{{2960}}{8} = 370\)More fragment bit is 0 because this is last fragment.Conclusions:IP datagram contains both data and header so 1424+20=1424.MF of third datagram is 0 and offset of third datagram is 370. Therefore Option 1 is the correct answer.