We can find the final velocity v at the end of the station.
v^2 = (v0)^2 + 2ax
v = sqrt{ (v0)^2 + 2ax }
v = sqrt{ (22.2 m/s)^2 - (2)(0.142 m/s^2)(206 m) }
v = 20.84 m/s
We can find the time to decelerate to this velocity.
v = v0 + at
t = (v - v0) / a
t = (20.84 m/s - 22.2 m/s) / -0.142 m/s^2
t = 9.58 seconds
The nose of the train is in the station for 9.58 seconds.