1. Equation of the surface

   The vertical cross-section is $y=\frac{x^2}{4}.$

   A block placed at some point on this curve will remain at rest only if the component of gravity along the tangent does not exceed limiting friction.

2. Condition for no slipping

   Let $\theta$ be the angle the tangent makes with the horizontal at the point of placement.

   For equilibrium on a rough incline: $mg\sin\theta \le \mu mg\cos\theta$ $\tan\theta \le \mu$

   Given: $\mu=0.5=\frac{1}{2}.$

   So the limiting condition is $\tan\theta = \frac{1}{2}.$

3. Slope of the curve

   For the curve $y=\frac{x^2}{4},$ the slope is $\frac{dy}{dx}=\frac{x}{2}.$

   But slope of the tangent is also $\tan\theta = \frac{dy}{dx}.$

   Therefore, at limiting equilibrium, $\frac{x}{2} = \frac{1}{2}$ which gives $x=1.$

4. Find the corresponding height

   Substitute $x=1$ into the curve equation: $y=\frac{x^2}{4}=\frac{1^2}{4}=\frac{1}{4}.$

   Hence the maximum height above the ground is $\boxed{\frac{1}{4}\text{ m}}.$

5. Check options

   • A: $\frac{1}{2}$ m ❌
   • B: $\frac{1}{3}$ m ❌
   • C: $\frac{1}{6}$ m ❌
   • D: $\frac{1}{4}$ m ✅

Therefore, the correct option is D.