Frame size $L_{\text {frame }}=2000 ~\text{bits}$
Sender data rate $R_{\text {data }}=500 \mathrm{Kbps}=500,000 \mathrm{bps}$
ACK size $L_{\text {ack }}=20 ~\text{bytes} = 160 ~\text{bits}$
ACK transmission rate $R_{\text {ack }}=100 \mathrm{Kbps}=100,000 \mathrm{bps}$
One-way propagation delay $T_p=50 \mathrm{~ms}=0.05 \mathrm{~s}$
Sender window size $W=5$ frames
$$
\begin{aligned}
T_{\text {tx-data }} & =\frac{L_{\text {frame }}}{R_{\text {data }}}=\frac{2000}{500000}=0.004 \mathrm{~s} \\\\
T_{\text {tx-ack }} & =\frac{L_{\text {ack }}}{R_{\text {ack }}}=\frac{160}{100000}=0.0016 \mathrm{~s}
\end{aligned}
$$
First frame sent at $t=0$, last bit at $t=0.004$.
It arrives at receiver at $t=0.004+T_p=0.054 \mathrm{~s}$.
Receiver sends ACK: transmission time of ACK $=0.0016 \mathrm{~s}$, so ACK last bit sent at $t=0.054+ 0.0016=0.0556$.
ACK arrives at sender at $t=0.0556+T_p=0.1056 \mathrm{~s}$.
So:
$$
\begin{gathered}
T_{\text {firstACK }}=0.1056 \mathrm{~s} \\\\
\mathrm{BDP}(\text { frames })=\frac{T_{\text {firstACK }}}{T_{\mathrm{tx} \text {-data }}}=\frac{0.1056}{0.004}=26.4 \text { frames }
\end{gathered}
$$
Our window size $W=5$ is less than $\mathbf{26.4}$, so throughput is limited by window size.
$$
\begin{aligned}
\text { Throughput } & =\frac{W \times L_{\text {frame }}}{T_{\text {firstACK }}}=\frac{5 \times 2000}{0.1056} \mathrm{bps} \\\\
= & \frac{10000}{0.1056} \approx 94696.97 \mathrm{bps} \\\\
& \approx 94.697 \mathrm{Kbps}
\end{aligned}
$$
