Friction

The frictional force decreases going up the slope. The highest point with an inclination of $\theta$ with the horizontal where the block of mass $m$ can be placed without slipping is when the downward force is balanced by the frictional force as follows:

$\quad mg\sin {\theta} = \mu_s mg \cos{\theta}\quad \Rightarrow \tan {\theta} = \mu_s = 0.5$

Since, $\space \tan {\theta} = \dfrac {dy}{dx} = \dfrac {2x}{20} = \dfrac {x}{10} \space$ .

And when $\tan{\theta} = 0.5\quad \Rightarrow \dfrac {x}{10} = 0.5$

$\quad \Rightarrow x = 5\quad \Rightarrow y = \dfrac {25}{20} = \boxed{1.25}$

Since co-efficient of friction is 0.5 hence tan(theta(max))=1/2. Now tan(theta)=dy/dx=x/10.Now x/10<1/2 hence X(max)=5. So y(max)=25/20=1.25