Answer
A torque of $~1.48~N\cdot m~$ must be applied.
Work Step by Step
We can find the rotational inertia of the system about this axis:
$I = \sum M_i~R_i^2$
$I = (\sum M_i)~(0.375~m)^2$
$I = (4.0~kg+3.0~kg+5.0~kg+2.0~kg)~(0.375~m)^2$
$I = 1.97~kg~m^2$
We can find the required torque:
$\tau = I~\alpha$
$\tau = (1.97~kg~m^2)(0.75~rad/s^2)$
$\tau = 1.48~N\cdot m$
A torque of $~1.48~N\cdot m~$ must be applied.
