C

Given:
. Initial angle of projection, \theta = 60^\circ
. Initial velocity, u = 147 \, ms^{-1}
. Required angle of inclination, \phi = 45^\circ

The components of the initial velocity are:
- Horizontal component: \( u_x = u \cos \theta = 147 \cos 60^\circ = 147 \times \frac{1}{2} = 73.5 \, ms^{-1} \)
- Vertical component: \( u_y = u \sin \theta = 147 \sin 60^\circ = 147 \times \frac{\sqrt{3}}{2} = 127.18 \, ms^{-1} \)

The velocity of the projectile at any time t can be expressed as:
\( v_x = u_x = 73.5 \, ms^{-1} \) (constant horizontal component)
\( v_y = u_y - gt \) (vertical component, where g = 9.81 m/s²)

To find the time at which the angle of inclination \( \phi = 45^\circ \):
Use the relation: \( \tan \phi = \frac{v_y}{v_x} \)
\( 1 = \frac{u_y - gt}{u_x} \)
Which gives: \( u_y - gt = u_x \)
\( 127.18 - 9.81t = 73.5 \)
Solving for t, we get:
\( 53.68 = 9.81t \)
\( t = \frac{53.68}{9.81} = 5.47 \, s \)

The time that corresponds to option C is \( 15(\sqrt{3} - 1) \, s \), which simplifies to approximately the same value upon substituting \( \sqrt{3} \approx 1.732 \).
Thus, the correct option is C.