Let s be the distance between the spaceship and the ground station.
Let the speed of signal be u, which is equal to 3 x 10^8 m/s.
The time (t) is 5 min, when converted into seconds, is 300 sec (5 min x 60 sec)
By the 3rd equation of motion…
s = ut + 1/2 (at^2) [here a is acceleration]
Since a is 0, the formula becomes…
s = ut + 0
(Distace = velocity ×time)
s = 3 x 10^8 ×300
s = 9 x 10^10 m or s = 9 x 10^7 km
So, the distance between the spaceship and the ground station is 9 x 10^10 m or 9 x 10^7 Km.