A park features a radius of 10 m . Assuming the road encircling it...

Given speed of the scooter, $v = 18 \text{ km/hr} = 18 \times \dfrac{5}{18} \text{ m/s} = 5 \text{ m/s}$

Radius of the circular park, $r = 10 \text{ m}$

For a vehicle to safely negotiate an unbanked circular curve, the required centripetal force is provided by the frictional force between the tyres and the road.

$\dfrac{mv^2}{r} \le \mu mg$

$\mu \ge \dfrac{v^2}{rg}$

Taking $g = 10 \text{ m/s}^2$, we get:

\mu_\min = \dfrac{5^2}{10 \times 10} = \dfrac{25}{100} = 0.25