Explanation

(c) Angle turned in three second, $\theta_{3s} = 2\pi \times 10 = 20\pi\ rad$
From $\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} \Rightarrow 20\pi = 0 + \frac{1}{2}\alpha \times (3)^{2}$
$\Rightarrow \alpha = \frac{40\pi}{9}rad/s^{2}$
Now angle turned in 6 sec from the starting
$\theta_{6s} = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 0 + \frac{1}{2} \times \left( \frac{40\pi}{9} \right) \times (6)^{2} = 80\pi\ rad$
$\therefore$ angle turned between $t = 3s$ to $t = 6s$
$\theta_{last\ 3s} = \theta_{6s} - \theta_{3s} = 80\pi - 20\pi = 60\pi$
Number of revolution $= \frac{60\pi}{2\pi} = 30\ rev$