Given $T_{processing}=0,Transmission_{ack}=0$
My total RTT here is $T_{data}+T_{prop}+T_{prop}=T_t+2*T_p$
My useful time will only be $N*T_{data}$ where $T_{data}$ will be transmission time for one frame.
so my efficiency
$\frac{N*T_{data}}{RTT}=\frac{T_{data}*N}{T_{data}+2*T_{prop}}=\frac{N}{1+2*\frac{T_p}{T_t}}=\frac{N}{1+2a}$
so, my efficiency would be 100% only if I send 1+2a packets into network in RTT time.
This 1+2a comes to be 201.
Now, this is the sender's window size. Receiver window size would be 1. So, the total you would need 202 distinct sequence numbers for frames and this requires $\lceil log_2202 \rceil = 8$ bits