Given , total length of packet = 1500 B
header length = 10 * 4 = 40 B [ As HLEN field value = 10 ]
So payload (data) length = 1500 - 40 = 1460 B
Now fragmentation is done into fragment sizes of 200 B [ As MTU of router given = 200 B ]
We should know that the header of packet is there in each fragment with some changes e.g. in flags and offset fields..
So payload length in each fragment = 200 - 40 = 160 B
Hence number of offsets in each fragment = 160 / 8 = 20
Now as 160 is a multiple of 8 , hence it is allowed as it is else we have to take nearest multiple of 8 which is less than 160..To accomodate this , we do padding in last fragment..Payload in each fragment should be a multiple of 8 because 1 offset = 8 bytes by convention..
So number of fragments needed = ceil ( 1460 / 160 ) = 10
Hence offset of last fragment = Offset of 1st fragment + (number of fragments - 1) * (no of offsets in each fragment)
= 100 + 9 * 20
Hence offset of last fragment = 280