C

To determine the time at which the projectile makes an inclination of 45° with the horizontal, we can use the concept of projectile motion.
Step 1: The initial velocity components can be calculated using the angle of projection. Given the angle of projection (θ) and the initial velocity (v), the components can be determined as follows:
$v_x = v imes ext{cos}( heta)$
$v_y = v imes ext{sin}( heta)$
Step 2: For a projectile, the angle of inclination with the horizontal at any time 't' can be given by:
$tan( heta) = \frac{v_y(t)}{v_x}$
Step 3: At 45°, $tan(45°) = 1$, hence we set:
$\frac{v_y(t)}{v_x} = 1$
This implies that $v_y(t) = v_x$
Step 4: Solving for the time 't', we get:
$v imes ext{sin}( heta) - g imes t = v imes ext{cos}( heta)$
Rearrange to find 't': $t = \frac{v \times (\text{sin}( heta) - \text{cos}( heta))}{g}$
Step 5: Plugging in the values, we determine that this simplifies to $15(\sqrt{3} - 1)$. Therefore, the correct answer is option C.