SInce sender Payload size=1600;
MTU of first network=1500Byte
No. Of fragments=1600/1480=1.089=~2
First fragment=1480(Payload)+20(Header)
Second fragment=120(payload)+20(Header)
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Now In second case::
MTU=480
Number of fragment=1480/460=~4
So
first fragment=456 +20,4560+20,4560+20,112+20
Total=4
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Total Data received by B=1700Byte