Here the altitude of the satellite is 36,000km, i.e. $d=36000 \times 1000$ m
speed of the signal $(v) = 3\times 10^8 m/s$
Propagation delay $T_p =\frac{2 \times 36000 \times 1000 }{3 \times 10^8} =0.24 s.$
(Packet must propagate to the satellite and then back)
Let packet size = L
bandwidth of channel B= 10Mbps
then transmission time $T_t =L/(10\times 10^6)$
$\text{Efficiency} = \frac{\text{Data transmitted}}{\text{Maximum data that could be transmitted}}$
In Go-Back-N, before first ACK comes, $N$ packets can be sent. ACK comes back after $T_t + T_p + T_p +T_{ack}$. $T_{ack} = 0$ as per question. So, we consider this time and during $T_t$ a packet of size $L$ can be sent. So,
$\text{Efficiency} = \frac{N \times L} { L + 2T_p \times B}$
$0.5 = \frac{100L}{L + 2\times 0.24 \times 10 \times 10^6}$
$0.5L + 2400000 = 100L$
$L = 2400000/99.5 = 24120.6 \text{ bits} \approx 3 \text{ Kbytes}$