@Bikram Sir, it is given in the question that:-
Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time.
thats why we are considering RTT as TT of data + PT of data (A to B) + PT of ack(B to A).
Secondly, We are getting 450 msec as RTT, I think it is because, it is given in the question that
The send and receive window sizes are 5 packets each
which shows that they are using Select and Repeat ARQ, and in This ARQ the receiver can send cumulative ack for all the data packets. i.e. I think sender is sending all five packets in one go itself and then waiting for ack to come in, That;s why it require TT of data, and in this TT it sends all 5 packets, and PT of ACk(B to A) in this time Sender is waiting for cumulative ack.
That's why the time is 50 + 2*200 = 450, ryt sir,