during the acceleration the time taken to reach full speed is 6 s
so angular acceleration will be given as
[tex]\alpha = \frac{\omega}{6}[/tex]
now from the same maximum speed it will stop in time t = 9.2 s
so deceleration is given as
[tex]\alpha' = \frac{\omega}{9.2}[/tex]
now we can say initially the torque will be given as
[tex]\tau_1 = I *\frac{\omega}{6}[/tex]
[tex]\tau_2 = I*\frac{\omega}{9.2}[/tex]
now we have
[tex]\frac{\tau_1}{\tau_2} = 9.2/6 = 1.533[/tex]