By equilibrium of block perpendicular to the incline,  $N=mg\mathrm{cos\theta }$.  So kinetic friction is,  $f_{\mathrm{k}}=\mu N=\mu mg\mathrm{cos\theta }$. Work done by constant force is given as, $W=FS\mathrm{cos\theta }$. Let distance covered by the mass is $x$ before coming at rest.

So, $W_{\mathrm{g}}=mg\times x\times \cos \left(90°-\mathrm{\theta }\right)=mgx\mathrm{sin\theta }$

$\Rightarrow W_{\mathrm{N}}=Nxcos90°=0$
Work done by the variable force is given as  $W={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}F\mathrm{dx}$.
So, $W_{\mathrm{f}}={\int }_0^{\mathrm{x}}{f}_{\mathrm{k}}dx=-{\int }_0^{\mathrm{x}}\left(\frac{2}{3}x\mathit{mg}\mathrm{cos\theta }\right) \mathrm{dx}$

$\Rightarrow W_{\mathrm{f}}=-mg\mathrm{cos\theta }\left(\frac{x^2}{2}\right)\times \frac{2}{3}=-mg\mathrm{cos\theta }\left(\frac{x^2}{3}\right)$

By work-energy theorem

$W_{\mathrm{g}}+W_{\mathrm{N}}+W_{\mathrm{f}}=∆K\mathit{.}E\mathit{.}$
$mgx\mathrm{sin\theta }+0-mg\mathrm{cos\theta } \left(\frac{{\mathrm{x}}^2}{3}\right)=0$
$\Rightarrow \mathrm{x}=3\tan \theta =3\tan 30°=\sqrt{3} \mathrm{m}$