Given altitude of the satellite is 36504 km. Let $A$ and $B$ be the two ground stations. Let $d$ be the distance between the two ground stations.
By Trigonometry, we can say, $tan$ $45^{\circ}$ = $perpendicular / base$ $\implies$
$tan$ $45^{\circ}$ = $\frac{36504}{(d/2)}$ $\implies$ $d = 2*36504 = 73008 km$
Now, In $GBN$, Utilization = $\dfrac{n}{1+2a}$, where n is the window size and $\large a = \dfrac{\text{Propagation Time}}{\text{Transmission Time}}$
Given utilization = $25\%$ or $1/4$ and $n = 127$, $v = 3*10^8m/s$, $ packet size = L$, $Bandwidth = 1Mbps$
So, $1/4$ = $\frac{1}{1+ \frac{2*73008*10^3*10^6}{3*10^8*L}}$ $\implies$ $L = 960 bits$ = $120 B$