C

Step 1: Consider the equations of motion for projectile motion. The horizontal and vertical components of the initial velocity are:
$$u_x = u \cos(\theta)$$
$$u_y = u \sin(\theta)$$

Step 2: At the point where the particle makes an angle of \( \frac{\theta}{2} \) with the horizontal, we can determine its velocity components.
The absolute magnitude of the velocity at that point can be described as:
$$v = \sqrt{u_x^2 + (u_y - gt)^2}$$
and the angle with the horizontal gives us the relationship:
$$\tan\left(\frac{\theta}{2}\right) = \frac{v_y}{v_x}$$

Step 3: Using the definition of radius of curvature (R):
$$R = \frac{(v^2)}{g \cos(\phi)}$$
where \( \phi \) is the angle of the trajectory at that point.
Substituting the appropriate values and considering that sin and cos relationships hold true, we can derive:
$$R = \frac{u^2}{2g \cos(\frac{\theta}{2})}$$

Therefore, option C is the correct answer.