To ensure the wheel flanges are not compressed, the net force from the rails must be a purely normal force, $\vec{N}$. This means there is no horizontal friction or side-thrust from the rails. The horizontal component of the normal force must provide the exact centripetal force required for the circular motion.

Let the train have mass $m$ and the track be banked at an angle $\theta$. The forces acting on the train in the plane perpendicular to its motion are gravity, $F_g = mg$, acting vertically downwards, and the total normal force from the rails, $N$, acting perpendicular to the banked surface.

We resolve the forces into horizontal and vertical components. The horizontal component provides the centripetal force, $F_c = m \frac{v^2}{R}$, required to keep the train on the circular path of radius $R$ at speed $v$.

Applying Newton's second law in the horizontal and vertical directions: Horizontal:

$$N\sin\theta = m\frac{v^2}{R}$$

Vertical (no vertical acceleration, so forces are balanced):

$$N\cos\theta = mg$$

To find the required angle $\theta$, we divide the first equation by the second:

$$\frac{N\sin\theta}{N\cos\theta} = \frac{m v^2/R}{mg}$$ $$\tan\theta = \frac{v^2}{gR}$$

Let $h$ be the height difference between the outer and inner rails and $d$ be the gauge of the railway. For a small banking angle $\theta$, the geometry of the rails gives the relation:

$$\tan\theta \approx \frac{h}{d}$$

Equating the two expressions for $\tan\theta$:

$$\frac{h}{d} = \frac{v^2}{gR}$$

Solving for the height difference $h$:

$$h = d \frac{v^2}{gR}$$

Now, we substitute the given values:

$d = 1435 \text{ mm} = 1.435 \text{ m}$ $R = 300 \text{ m}$ $v = 20 \text{ m/s}$

Let's use $g = 9.8 \text{ m/s}^2$.

$$h = (1.435 \text{ m}) \frac{(20 \text{ m/s})^2}{(9.8 \text{ m/s}^2)(300 \text{ m})}$$ $$h = 1.435 \frac{400}{2940} \text{ m}$$ $$h \approx 0.1952 \text{ m}$$

Converting to millimeters for a more practical unit:

$h \approx 195.2 \text{ mm}$