At the base of a smooth inclined plane with an inclination of 30^ ^...

For particle B to be in equilibrium, the net force acting on it along the inclined plane must be zero.

The downward component of the weight of particle B along the incline is balanced by the upward electrostatic force of repulsion from particle A.

$mg \sin \theta = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$

Substituting the given values $m = 0.1 \text{ kg}$, $g = 9.8 \text{ m/s}^2$, \theta = 30^{}^{{}^{\circ}}, and $q = 2.0 \times 10^{-6} \text{ C}$:

0.1 \times 9.8 \times \sin 30^{}^{{}^{\circ}} = \dfrac{9 \times 10^9 \times (2.0 \times 10^{-6})^2}{r^2}

$0.1 \times 9.8 \times 0.5 = \dfrac{9 \times 10^9 \times 4.0 \times 10^{-12}}{r^2}$

$0.49 = \dfrac{0.036}{r^2}$

$r^2 = \dfrac{0.036}{0.49} \approx 0.0735 \text{ m}^2$

$r \approx 0.271 \text{ m} = 27.1 \text{ cm}$

The closest value among the options is $27 \text{ cm}$.