Let the sender window size be $N$.
$\text{One way propagation delay} =100\,ms=0.1\,s$
$\text{Transmission delay}_{packet}=\frac{\text{Size of data frame}}{\text{Link bit rate}}=\frac{2000}{10^{6}}=0.002\,s$
$\text{Transmission delay}_{ACK}=\frac{\text{Size of ACK frame}}{\text{Link bit rate}}=\frac{10}{10^{6}}=0.00001\,s$
$\text{Link Utilization}(\eta )=\frac{\text{Useful Data Transfer time}}{\text{Total time}}$
$\implies \eta = \frac{N.(T_{f})}{T_{f}+2(T_{p})+T_{ACK}}$
$\implies N = \left \lceil \frac{\eta.\left(T_{f}+2(T_{p})+T_{ACK}\right)}{T_{f}}\right \rceil = \left \lceil \frac{0.5(0.002+0.2+0.00001)}{0.002}\right \rceil =\lceil 50.5025 \rceil =51$
Correct Answer: $51$