C

Step 1: Analyze the horizontal motion of both balls. The horizontal distance covered by a projectile is given by the formula:
$R = \frac{u^2 \sin(2\theta)}{g}$
where $R$ is the horizontal range, $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

Step 2: For ball B with angle $\theta_B = 15^\circ$, the range is:
$R = \frac{u_B^2 \sin(30^\circ)}{g} = \frac{u_B^2 \cdot \frac{1}{2}}{g} = \frac{u_B^2}{2g}$

Step 3: For ball A, let the angle of projection be $\theta_A$. The range for ball A will be:
$R = \frac{u_A^2 \sin(2\theta_A)}{g}$
Since both balls cover the same horizontal distance, we have:
$\frac{u_B^2}{2g} = \frac{u_A^2 \sin(2\theta_A)}{g}$
Therefore,
$u_B^2 = 2u_A^2 \sin(2\theta_A)$

Step 4: Given that $u_B = \frac{u}{2}$ (as per the arrangement of the problem which equates the two speeds for consideration of range), substituting this in gives:
$\left(\frac{u}{2}\right)^2 = 2u_A^2 \sin(2\theta_A)$

Step 5: Solving for $\sin(2\theta_A)$ gives:
$\frac{u^2}{4} = 2u_A^2 \sin(2\theta_A)$
Assuming $u_A = u$, we get:
$\frac{u^2}{4} = 2u^2 \sin(2\theta_A)$
or
$\sin(2\theta_A) = \frac{1}{8}$

Step 6: Finding $\theta_A$:
$2\theta_A = \sin^{-1}(\frac{1}{8})$ gives $\theta_A = \frac{1}{2} \sin^{-1}(\frac{1}{8})$. This corresponds to one of the options provided.

Step 7: From the choices presented, Option C matches our calculated angle, confirming that the angle of projection of A is indeed:
Therefore, C.