We know that the rod has length l and is free to rotate. Also, it makes an angle of `45^o` with the horizontal.

First, we note that `tau = I times omega` , where `tau` is the torque, `I` the moment of inertia and `w` the angular acceleration.` `

We are given the moment of inertia: `I = (ml^2)/3` and we want to look for the angular acceleration. To do so, we first need to solve for the torque acting on the rod. Torque is given by the following:` `

`tau = (mgl)/2 cos(theta)` , which we can express, by substituting the angle, as:

`tau = (mgl)/2 cos(45^o) = (mgl)/2 * (sqrt(2)/2) = (sqrt(2)mgl)/4`

Now that we have an expression for the torque, we solve for the angular acceleration:

`tau = I times w rArr w = tau / I`

`omega = [(sqrt(2)mgl)/4]/[(ml^2)/3]`

`omega = (3sqrt(2)g)/(4l)`

Note that this is equivalent to:

`omega = (3g)/(2sqrt(2)l)`

[To get the transformation, simply multiply either expression by `sqrt(2)/sqrt(2)` to get the other.]` `

Hence, answer is A