In the above diagram all surfaces are frictionless. The horizontal force has to be applied on wedge such that equilibrium steady state spring is compressed by $\frac{\mathrm{mgsin\theta }}{\mathrm{K}}$ is $\text{xmgTanθ}$ then $\mathrm{x}$ value is

If the wedge is moving with an acceleration 'a', from the frame of wedge

Applying newton’s law among incline

$\mathrm{kx}+\mathrm{mgsin\theta }=\mathrm{macos\theta }$

$\mathrm{mgsin\theta }+\mathrm{mgsin\theta }=\frac{\mathrm{F}}{2}\mathrm{cos\theta }$ (equilibrium)

$2\mathrm{mgsin\theta }=\frac{\mathrm{F}}{2}\mathrm{cos\theta }$

$\mathrm{F}=4\mathrm{mg}\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}$

$\mathrm{F}=4\mathrm{mg}\mathrm{Tan\theta }$

$\therefore \mathrm{x}=4$