A

Step 1: Convert the speed from km/hr to m/s.
To convert 18 km/hr to m/s, use the conversion factor:
$$ v = 18 \times \frac{1000}{3600} = 5 \, m/s $$
Step 2: Determine the centripetal acceleration.
The centripetal acceleration ($$ a_c $$) is given by:
$$ a_c = \frac{v^2}{r} $$
where $$ v = 5 \, m/s $$ and $$ r = 10 \, m $$.
Thus,
$$ a_c = \frac{(5)^2}{10} = \frac{25}{10} = 2.5 \, m/s^2 $$
Step 3: Calculate the angle of banking.
The angle of banking ($$ \theta $$) can be found using the formula:
$$ \tan(\theta) = \frac{v^2}{rg} $$
where $$ g = 9.8 \, m/s^2 $$ is the acceleration due to gravity.
Thus,
$$ \tan(\theta) = \frac{(5)^2}{10 \times 9.8} = \frac{25}{98} \approx 0.2551 $$
Now, find the angle:
$$ \theta = \tan^{-1}(0.2551) \approx 14.74^\circ $$
Step 4: Minimum friction coefficient for no banking.
In a horizontal road without banking, the necessary frictional force must provide the required centripetal force.
The frictional force $$ f $$ is given by:
$$ f = \mu m g $$
Setting it equal to the required centripetal force:
$$ \mu m g = \frac{mv^2}{r} $$
Canceling $$ m $$ gives:
$$ \mu g = \frac{v^2}{r} $$
Now substituting the known values:
$$ \mu \cdot 9.8 = \frac{(5)^2}{10} $$
$$ \mu \cdot 9.8 = 2.5 $$
$$ \mu = \frac{2.5}{9.8} \approx 0.2551 $$
Conclusion:
Therefore, the angle of banking is approximately $$ 14.74^\circ $$, and the minimum friction coefficient required is approximately $$ 0.2551 $$. Hence, the correct option is A.