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

Given $v = 18 \text{ km/h} = 5 \text{ m/s}$, $m = 100 \text{ kg}$, $R = 100 \text{ m}$.

The track consists of two circular arcs, each subtending 90^{}^{{}^{\circ}} at their respective centres. By symmetry, the radius vector at point $C$ makes an angle \theta = 45^{}^{{}^{\circ}} with the vertical.

Just before crossing $C$, the cycle is on the convex part of the track. The centripetal force is provided by the component of weight towards the centre minus the normal reaction.

mg \cos 45^{}^{{}^{\circ}} - N_1 = \dfrac{mv^2}{R}

N_1 = mg \cos 45^{}^{{}^{\circ}} - \dfrac{mv^2}{R}

Substituting the values:

$N_1 = 100 \times 10 \times \dfrac{1}{\sqrt{2}} - \dfrac{100 \times (5)^2}{100}$

$N_1 = 500\sqrt{2} - 25 \approx 707 - 25 = 682 \text{ N}$

Just after crossing $C$, the cycle is on the concave part of the track. The centripetal force is provided by the normal reaction minus the component of weight away from the centre.

N_2 - mg \cos 45^{}^{{}^{\circ}} = \dfrac{mv^2}{R}

N_2 = mg \cos 45^{}^{{}^{\circ}} + \dfrac{mv^2}{R}

Substituting the values:

$N_2 = 100 \times 10 \times \dfrac{1}{\sqrt{2}} + \dfrac{100 \times (5)^2}{100}$

$N_2 = 500\sqrt{2} + 25 \approx 707 + 25 = 732 \text{ N}$