The forces acting on the block are:
1. The component of gravitational force along the incline, \( F_{\text{gravity}} = mg \sin \theta \)
2. The maximum static friction force, \( F_{\text{friction}} = \mu_s mg \cos \theta \)
For the block to slide down, the component of gravitational force must overcome the friction force.
\[
F_{\text{gravity}} = mg \sin \theta = 2 \times 9.8 \times \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8 \, \text{N}
\]
\[
F_{\text{friction}} = \mu_s mg \cos \theta = 0.4 \times 2 \times 9.8 \times \cos 30^\circ = 0.4 \times 2 \times 9.8 \times 0.866 = 6.77 \, \text{N}
\]
Since the frictional force is greater than the component of gravity along the incline, the block does not slide.
Final answer
Answer: \(\boxed{\text{B}}\)