Here the thing to keep in mind is :
For every byte in TCP , we have a sequence number assigned .
Hence for wraparound of all bytes possible , we need to cover maximum number of sequence numbers i.e. we have to start from sequence number 0 and continue till (232 - 1) as sequence number field in TCP is of 32 bits . After (232 - 1) , the numbering will begin from 0 again which is known as "wraparound of sequence numbers" .
Hence data involved before wraparound of sequence numbers = 232 B (as 1 sequence number is for 1 byte)
= 235 bits
Given data rate of the network = 200 Mbps
= 200 * 106 bps
Hence time taken for wraparound of sequence numbers = Data size / Data rate
= 235 / (200 * 106) s
= 171.799 s