Answer : $\large\frac{\tan \theta}{k}$
The net downward force on the body at a distance x is
$f(x) =mg \sin \theta - \mu mg \cos \theta$
$\qquad= mg( \sin \theta - \mu \cos \theta)$
$\qquad= mg (\sin \theta- kx \cos \theta)$
$\therefore f(x) =0$ at a value of $x =x_0$ given by
$\sin \theta -kx_0 \cos \theta=0$
Which gives $x_0 = \large\frac{\tan \theta}{k}$