A body of mass 2 kg is situated on a rough inclined plane of...

The component of the weight of the block acting down the inclined plane is:

W_{\parallel} = mg \sin 30^{}^{{}^{\circ}}

$W_{\parallel} = 2 \times 9.8 \times \dfrac{1}{2} = 9.8 \text{ N}$

The maximum static friction force acting up the incline is:

f_{\text\max} = \mu_s N = \mu_s mg \cos 30^{}^{{}^{\circ}}

f_{\text\max} = 0.2 \times 2 \times 9.8 \times \dfrac{\sqrt{3}}{2} \approx 3.39 \text{ N}

Since the downward driving force ($9.8 \text{ N}$) is strictly greater than the maximum static friction ($3.39 \text{ N}$), the block will slide down the incline under its own weight.

Therefore, no additional external force is required to make the block move down the incline. The required force is zero.