B

Step 1: Convert the speed from km/h to m/s.
The speed given is 36 km/h. To convert it to m/s, use the conversion factor:
$$ 1 ext{ km/h} = \frac{1}{3.6} ext{ m/s} $$
Hence,
$$ 36 ext{ km/h} = \frac{36}{3.6} = 10 ext{ m/s} $$
Step 2: Identify the forces acting on the vehicle.
When a vehicle is on a banked turn, the forces acting on it are:
1. The gravitational force (Weight) acting downwards: $$ F_g = mg $$
2. The normal force (N) acting perpendicular to the surface of the bank.
3. The frictional force (f) which can act up or down the bank to prevent slipping.
Step 3: Set up the equations of motion.
The forces must balance in both vertical and horizontal directions.
For circular motion, we have:
$$ \frac{mv^2}{r} = N \sin(\theta) + f \cos(\theta) $$
and
$$ mg = N \cos(\theta) - f \sin(\theta) $$
Step 4: Express normal force and friction.
The maximum static friction can be expressed as:
$$ f_{max} = \mu N $$
where $$ \mu = 0.4 $$ is the coefficient of static friction.
Step 5: Use the given radius and speed to find the angle of banking.
We can't compute the exact \\theta without additional data, but we use the equation of banking for ideal circumstances (no friction), which is:
$$ \tan(\theta) = \frac{v^2}{rg} $$
where $$ g \approx 9.81 \text{ m/s}^2 $$, $$ r = 20 ext{ m} $$, and $$ v = 10 ext{ m/s}. $$
Therefore,
$$ \tan(\theta) = \frac{10^2}{20 \cdot 9.81} \approx 0.5102 $$
Step 6: Find the minimum and maximum possible speeds.
The maximum speed considering friction acting up the bank would be given by:
$$ v_{max} = \sqrt{r g (\tan(\theta) + \mu)} $$
And the minimum speed considering friction acting down the bank would be:
$$ v_{min} = \sqrt{r g (\tan(\theta) - \mu)} $$
Using the values, we can calculate these speeds. After computation:
$$ v_{max} \approx 13.53 \text{ m/s} $$
and
$$ v_{min} \approx 6.2 \text{ m/s} $$
Step 7: Conclusion.
Therefore, the possible speeds so that the vehicle neither slips down nor skids up are between 6.2 m/s and 13.53 m/s, which includes the speed 10 m/s. Therefore, the answer choice corresponding to this range is option B.