$\theta = {\omega _1}t + \frac{1}{2}\alpha \,{t^2}$

${\theta _1} = 0 + \frac{1}{2}\alpha \,{(2)^2} = 2\alpha$ …..(i)  $ [ {\omega _1} = 0, \,t = 2\,\sec, \,\theta = {\theta _1}]$

${\theta _1} + {\theta _2} = 0 + \frac{1}{2}\alpha \,{(4)^2}$          $[{\omega _1} = 0, \,t = 2 + 2 = 4\,\sec , \,\theta = {\theta _1} + {\theta _2}]$

${\theta _1} + {\theta _2} = 8\alpha $….(ii)

${\theta _1} = 2\alpha, \,{\theta _2} = 6\alpha $

$\therefore \,\,\frac{{{\theta _2}}}{{{\theta _1}}} = 3$