\({F_l} = 0.7 \times 2 \times 10 \times \cos 30^\circ = 12\;N\)(approximately)
But when the block is lying on the inclined plane then component of weight down the plane \( = mg\sin \theta \) \( = 2 \times 9.8 \times \sin 30^\circ = 9.8\;N\)
It means the body is stationary, so static friction will work on it
\(\therefore \) Static friction \(=\) Applied force \(= 9.8 \,N\)
$\left[g=10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$