From the Force Polygon
$\dfrac{R_A}{\sin 45^\circ} = \dfrac{20 + 10}{\sin 105^\circ}$
$R_A = 21.96 \, \text{ kN}$
From the Free Body Diagram
$\Sigma M_B = 0$
$4(R_A \cos 30^\circ) = 20(3) + 10x$
$4(21.96 \cos 30^\circ) = 20(3) + 10x$
$10x = 16.072$
$x = 1.61 \, \text{ m}$ answer