Given:

The equation of the surface: $$y = \frac{x^2}{4}$$.

Coefficient of friction: $$\mu = 0.5$$.

Gravitational acceleration: $$g$$ (assumed constant).

1. Slope of the Surface:

The slope of the surface at any point is given by:

$$ \tan\theta = \frac{dy}{dx} $$

From the equation of the surface:

$$ y = \frac{x^2}{4} $$

Differentiating with respect to $x$:

$$ \frac{dy}{dx} = \frac{x}{2} $$

Thus, the slope at any point is:

$$ \tan\theta = \frac{x}{2} $$

2. Forces Acting on the Block:

At the point where the block is placed:

Weight of the block acts vertically downward: $ mg $.

Normal force acts perpendicular to the surface.

Frictional force acts parallel to the surface, opposing the component of the weight that causes slipping.

3. Condition for No Slipping:

The block will not slip if the frictional force is sufficient to counteract the component of the gravitational force parallel to the slope. The frictional force is:

$$ f = \mu N $$

The normal force $N$ is given by:

$$ N = mg \cos\theta $$

The component of weight parallel to the slope is:

$$ F_{\text{parallel}} = mg \sin\theta $$

For the block to not slip:

$$ f \geq F_{\text{parallel}} $$

Substitute $f = \mu N$ and $N = mg \cos\theta$:

$$ \mu (mg \cos\theta) \geq mg \sin\theta $$

Simplify:

$$ \mu \cos\theta \geq \sin\theta $$

Divide through by $\cos\theta$:

$$ \mu \geq \tan\theta $$

4. Maximum Slope Without Slipping:

From the above condition:

$$ \tan\theta \leq \mu $$

Substitute $\mu = 0.5$:

$$ \tan\theta \leq 0.5 $$

Thus:

$$ \frac{x}{2} \leq 0.5 $$

$$ x \leq 1 $$

5. Maximum Height:

The height $y$ of the block at $x = 1$ is obtained from the surface equation:

$$ y = \frac{x^2}{4} $$

Substitute $x = 1$:

$$ y = \frac{1^2}{4} = \frac{1}{4} \, \text{m} $$

Final Answer:

The maximum height above the ground at which the block can be placed without slipping is:

$$ \boxed{\frac{1}{4} \, \text{m}} $$

Thus, the correct option is Option D.