A

To find the angle of elevation that produces the maximum range for a projectile, we can use the formula for range in projectile motion, which is given by:
R = \frac{v^2 \sin(2\theta)}{g}
where R is the range, v is the initial velocity (muzzle speed), \theta is the angle of elevation, and g is the acceleration due to gravity.
To maximize the range, we need to maximize \sin(2\theta). The maximum value of \sin(2\theta) is 1, which occurs when 2\theta = 90^{\circ} or \theta = 45^{\circ}. Therefore, the angle of elevation of the gun that achieves this range is:
\theta = 45^{\circ}
Hence, the correct answer is A.