Number of fragments =ceil(4404-20/1500-2)=3(approx) Take integer
First Fragments.
Offset of first fragments :: 0 and MF =1
Second Fragments:
Offset of IInd fragments :: 185 and MF =1
Third Fragments::
Offset of IIIrd fragments :: 370 and MF =1
And datagram length is nothing but total length=1424+20=1444 Byte
Therefore option A will be right.