Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts.
When the data is transmitted using a duplex link, Round Trip Time (RTT) do not include the time taken by ACK frame to reach sender from the receiver. This is because the ACK frames are piggybacked with the data frames that are coming from the receiver for the sender (maybe a bit later).
Aforementioned, we must now calculate the number of frames transmitted during propagation time (rather than round trip time)
$No.\ of\ frames = \left(25*10^{-3}*10^{6}\right)/10^3 = 25$
Thus, we will need minimum of $5\ bits$ to represent sequence number distinctly.
HTH
