If you traveled this distance at a constant speed - the speed would have to be 300m/60s = 5 m/s. To travel the same distance with linear acceleration from stand-still and using the same time, speed at the end would have to be twice that of the constant speed example.
The generic formula for this type of problem is then as follows:
d = vt + (1/2)a(t^2)
d = distance
v = initial speed
t = time
a = acceleration
With initial speed at 0 and solving for "a" you get:
a = d / ((1/2)(t^2))
a = 300 / ((1/2)(60^2)) = 0.17