以36km/hr的速度在半徑200m的彎路上行駛之火車

36 km/h = 10 m/s

Let d be the difference in height between the inner and outer rails.
Hence, tan(a) = d/1.2
where angle a is the angle subtended by the outer rail at the inner rail
[please check the given firgure for the separation between the rails, 120 m is unrealistic, I presume that it is 120 cm, or 1.2 m]

Let R1 and R2 be the normal reaction on the inner wheels and outer wheels of a train car.
Hence, (R1+R2).sin(a) = m(10^2)/200
and (R1+R2).cos(a) = mg
where m is the mass of a train car

Dividing, tan(a) = 10^2/200g = 0.05
thus, d/1.2 = 0.05
d = 1.2 x 0.05 m = 0.06 m = 6 cm