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As given in the question MTU size is 200 at router which contains 20 bytes of IP header

Total data bytes - 600

1st fragment - 

20

176

total size of of fragment should be divisible by 8(concept called scaling)

offset - 0

2nd fragment - 

20

176

offset - 0 + 176(data bytes of first fragment)/8 =22

3rd fragment - 

20

176

offset - 22 + 176(data bytes of second fragment)/8 =44

4th fragment - 

20

72

offset - 44 + 176(data bytes of third fragment)/8 = 66

Answer 1

For 1st question answer is d and for 2nd one b