Protocol is Selective Repeat.
Sequence bits $(k)=7$.
Total sequence space $=2^k=2^7=128$.
The rule for SR is $W_S+W_R \leq 2^k$.
Given $W_S=W_R$ and the space is fully utilized, $W_S+W_S=128$.
$$2 \times W_S=128 \Longrightarrow W_S=\mathbf{6 4}$$
$$W=64 .$$
$\eta$ (Utilization) $=80 \%=0.8$
$L$ (Frame Size) $=1000$ Bytes $=8,000$ bits
$T_p$ (One-way) $=25 \mathrm{msec}=0.025$ seconds
RTT $=2 \times T_p=0.050$ seconds
$$
T_t=L / B=8,000 / B
$$
\begin{aligned}
& \eta=\left(W \times T_t\right) /\left(T_t+R T T\right) \\\\
& 0.8=(64 \times(8000 / B)) /((8000 / B)+0.05) \\\\
& 0.8 \times((8000 / B)+0.05)=64 \times(8000 / B) \\\\
& (0.8 \times 8000 / B)+(0.8 \times 0.05)=512,000 / B \\\\
& (6400 / B)+0.04=512,000 / B \\\\
& 0.04=(512,000 / B)-(6400 / B) \\\\
& 0.04=(512,000-6400) / B \\\\
& 0.04=505,600 / B \\\\
& B=505,600 / 0.04 \\\\
& B=12,640,000 \text { bits/second } \\\\
& B=12,640,000 / 10^6 \mathrm{Mbps}=\mathbf{1 2 . 6 4} \mathrm{Mbps}
\end{aligned}
This rounds to $12.6$ $\mathrm{ Mbps}$.
