Header Fields of the Third IP Fragment (MTU 1500)

Given: MTU = 1500 bytes, received IP packet = 4404 bytes, IP header = 20 bytes.

Step 1: Data that must be fragmented

$$\text{Data} = 4404 - 20 = 4384\ \text{bytes}$$

Step 2: Maximum data per fragment

$$\text{Max data} = \text{MTU} - \text{header} = 1500 - 20 = 1480\ \text{bytes}$$

The offset is counted in 8-byte blocks, so every non-final fragment must carry a multiple of 8 bytes: $1480 \bmod 8 = 0$, so 1480 is legal and is used as the data size of fragments 1 and 2.

Step 3: Fragmentation plan

Fragment	Data bytes (original data field)	Data size	Total length	Offset (value = bytes/8)	M flag
1st	0 - 1479	1480	$1480 + 20 = 1500$	$0/8 = 0$	1
2nd	1480 - 2959	1480	$1480 + 20 = 1500$	$1480/8 = 185$	1
3rd	2960 - 4383	$4384 - 2960 = 1424$	$1424 + 20 = 1444$	$2960/8 = 370$	0

Step 4: Fields in the header of the third fragment

Field	Value
Total Length	1444 bytes (1424 data + 20 header)
Header Length (HLEN)	5 (i.e. 20 bytes)
Identification	same as the original packet
Flags	MF = 0, DF = 0 (this is the last fragment)
Fragment Offset	370 (means $370 \times 8 = 2960$ bytes of data precede this fragment)
Protocol	copied unchanged from the original header

Final answer: Total Length = 1444 bytes, MF flag = 0, Fragment Offset = 370 (header length stays 20 bytes, identification unchanged).