$m =2\,kg$

$\theta=30^{\circ}$

$\mu=0.5$

$g =10\,m / s ^2$

The contact force is $F _{ i }$ is the resultant of frictional force and normal force.

From the free body diagram,

Normal force, $N = mg \cos \theta$

$N=2 \times 10 \times \cos 30^{\circ}$

$N=17.320\,N$

The friction force, $F _{ r }=\mu \mu g \cos \theta$

$F_r=0.5 \times 2 \times \times 10 \times \cos 30^{\circ}$

$F _{ r }=8.660\,N$

$N$ and $F_r$ are perpendicular to each other.

The contact force is, $F_c=\sqrt{N^2+F_r^2}$

$F_c=\sqrt{(17.320)^2+(8.699)^2}$

$F _{ c }=19.381\,N$

We can say approximately $20\,N$