See total size of message is 4500 now it will pass through first network which has MTU of 2500 so first fragment will be 2480 bytes of data and rest 20 bytes will be header and next fragment will be 2020 bytes of data and 20 bytes of header . now At network layer 2 whose MTU is 1500 bytes so first fragment which is of 2480 bytes data is fragmented into 1480 bytes data and 20 bytes header. second fragment it will be 1000 bytes of data and 20 bytes header so total of 2480 of first fragment of network 1 is complete. now coming to second fragment of 2020 bytes it will be split into 1480 bytes data and 20 bytes header so this is our third fragment and fourth fragment will be 540 bytes of data and 20 bytes header. now for 3 rd fragement data which is ahead of third fragment is 1480+1000 =2480 bytes so fragment offset of third one is 2480/8=310. So Our final answer is 310.
