Number of packet fragments = ⌈ (total size of packet)/(MTU) ⌉
= ⌈ 4404/1500 ⌉
= ⌈ 2.936 ⌉
= 3
So Datagram with data 4404 byte fragmented into 3 fragments.
The first frame carries bytes 0 to 1479 (because MTU is 1500 bytes and HLEN is 20 byte so the total bytes in fragments is maximum 1500-20=1480).
the offset for this datagram is 0/8 = 0.
The second fragment carries byte 1480 to 2959.
The offset for this datagram is 1480/8 = 185.
finally the third fragment carries byte 2960 to 4404.
the offset is 370.and for all fragments except last one the M bit is 1.
so in the third bit M is 0.