I did the first part (using parallel axis theorem) and showed that intertia.
The problem is in the second part, I know that $C=I\frac{d^2\theta}{dt^2}$ , where C is the moment. So in this case it will be moment about A.
So it's like I don't know how to find the velocity at C, I don't know how to do this part. Any help is very much appreciated.
Let's start with the simplest version of this problem. Suppose we take a massless rigid rod, attach a weight to one end, and run an axis through the other end. In that case the center of gravity is just at the end of the rod. Thus if the rod starts off horizontal and is released, the initial acceleration is just that of gravity ($g$) since the gravitational force $mg$ acts here directly.
On the other hand, suppose we have a rod (length $L$) with a uniformly-distributed mass $m$ (no weight on the end). Then the center of gravity is at the center of the rod, rather than at one end, so the force of gravity $mg$ produces a torque $\tau=(L/2)\cdot(mg)=mgL/2$. But the moment of inertia of this rod (relative to the chosen axis) is $I=\frac{1}{3}m L^2$. Thus Newton's 2nd law for torques implies an angular acceleration of $\alpha=\tau/I=3g/2L$. Since this angular acceleration is valid throughout the rod, it applies as well at the end of the rod; the relation between linear and angular acceleration then gives an initial linear acceleration of $a=L\alpha=3g/2$.
For the problem at hand, the moment of inertia $I$ is no longer so simple but can still be calculated (as you've already done.) Moreover, the object is sufficiently symmetric such that the center of gravity still lies halfway between the point of interest and the axis. Therefore, one can use the result from part $(a)$ to deduce the initial acceleration.