A track consists of two circular parts ABC and CDE of equal radius...

The track is in a vertical plane. For the two circular arcs to join smoothly and each subtend $90^\circ$ at their respective centers, the track must have horizontal tangents at the lowest point of the concave part (valley) and the highest point of the convex part (hill). The angle $\theta$ made by the radius with the vertical varies from $0^\circ$ to $45^\circ$ for both parts.

For the cyclist to move at a constant speed $v$, the net tangential force must be zero. Thus, the required friction force balances the tangential component of gravity:
$f = mg \sin \theta$

The normal force $N$ depends on whether the cyclist is on the concave or convex part of the track:
On the concave part (ABC): $N = mg \cos \theta + \dfrac{mv^2}{R}$
On the convex part (CDE): $N = mg \cos \theta - \dfrac{mv^2}{R}$

To prevent slipping, the coefficient of friction $\mu$ must satisfy $\mu \ge \dfrac{f}{N}$ everywhere on the track. The required $\mu$ will be maximum where $f$ is maximum and $N$ is minimum. This occurs on the convex part at the maximum angle, $\theta = 45^\circ$.

Given values:
$v = 18 \text{ km/h} = 5 \text{ m/s}$
$R = 100 \text{ m}$
$g = 10 \text{ m/s}^2$

Calculate the term $\dfrac{v^2}{Rg}$:
$\dfrac{v^2}{Rg} = \dfrac{5^2}{100 \times 10} = \dfrac{25}{1000} = 0.025$

Substitute the values into the condition for the convex part at $\theta = 45^\circ$:
\mu_\min = \dfrac{mg \sin 45^\circ}{mg \cos 45^\circ - \dfrac{mv^2}{R}} = \dfrac{\sin 45^\circ}{\cos 45^\circ - \dfrac{v^2}{Rg}}

\mu_\min = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} - 0.025} = \dfrac{1}{1 - 0.025\sqrt{2}}

Using $\sqrt{2} \approx 1.414$:
\mu_\min = \dfrac{1}{1 - 0.025 \times 1.414} = \dfrac{1}{1 - 0.03535} = \dfrac{1}{0.96465} \approx 1.037

Therefore, the minimum friction coefficient required is $1.037$.

Answer: $1.037$