An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are

Given:

• Total IP packet size = $4404$ bytes

• IP header size = $20$ bytes

• Payload data size = $4404 - 20 = 4384$ bytes

• MTU = $1500$ bytes

The maximum payload capacity per fragment is $1500 - 20 = 1480$ bytes (which is a multiple of $8$, as required for fragmentation offset: $1480 / 8 = 185$).

Let us trace the fragments:

1. Fragment 1:

   • Data bytes: $0$ to $1479$ ($1480$ bytes)

   • Datagram length = $1480 + 20 = 1500$ bytes

   • Fragment Offset = $0 / 8 = 0$

   • $\text{MF} = 1$

2. Fragment 2:

   • Data bytes: $1480$ to $2959$ ($1480$ bytes)

   • Datagram length = $1480 + 20 = 1500$ bytes

   • Fragment Offset = $1480 / 8 = 185$

   • $\text{MF} = 1$

3. Fragment 3:

   • Remaining data: $4384 - 2960 = 1424$ bytes (bytes $2960$ to $4383$)

   • Datagram length = $1424 + 20 = 1444$ bytes

   • Fragment Offset = $2960 / 8 = 370$

   • $\text{MF} = 0$ (last fragment)

Thus, the third fragment has MF bit: 0, Datagram Length: 1444, Offset: 370, which matches (A).