Why does bouyant force act at the centre of gravity? 
If the length of the rod immersed inside the liquid is $\frac{L/2}{\sin(\theta)}$, then will buoyant force act at $\frac{L/4}{\sin(\theta)}$? Can someone please explain this?
 A: Tl;dr
Buoyancy pushes up equally in all parts of the object of the same reason that gravity pulls equally in all parts. These two forces do not make the object rotate about itself, meaning their torques must balance out about some centre'ish point. This point is where you would draw the total force of buoyancy or of gravity, and this point happens to be the centre-of-geometry in this scenario.

More depth
Buoyancy actually pushes up at every single point on the downwards surface. Not just from one point. Just like gravity pulls down in every single particle in an object. Not just in the centre-of-gravity.
But it is not easy to work with force distributions. So when the distribution aspect is not relevant - when only the total effect of the force distribution as a whole is relevant, which it often is in Newtonian mechanics on rigid bodies - then we'd often simplify the force distribution to just a single force. We simply sum all small forces up and consider only the total force as was it one single force acting somewhere. No point actually feels this entire force - only the object as a whole feels it, but spread out. It is a model no matter what we do. But models can be useful in scenarios where the left-out details are not important.
The question is now from where to draw this total force. Which point in an object should we choose as a representative for the whole object. This leads us to the typical method in physics of modelling objects as point-masses, when their shape, size and extend isn't relevant. Which point can represent the whole object in this way?
In the case of gravity, we see that an object in free fall doesn't rotate. Clearly, any gravitational pull on the right side must be balanced out by a gravitational pull on the left side so that their torques cancel out. The point that these torques cancel out about is a good choice as a representative. We call it the centre-of-gravity. This point coincides with the centre-of-mass for human-scale not-too-tall objects (objects across which the gravitational pull is constant).
In the case of a body of water, the mass distribution is even, uniform. So the centre-of-gravity (or centre-of-mass) is in what we might call the centre-of-geometry. In the very middle. It is from this point that we can choose to draw the total gravitational pull (the weight).
Buoyancy is a result of gravity. Or rather gravitational differences. It is an upwards force that only exists due to a difference in (total) gravitational pull between the submerged object and the displaced water volume. When the object is less dense than the displaced water, then that water amount is pulled down more strongly than the object is, resulting in an apparent upwards force on the object.
The buoyancy force distribution is thus equivalent to the gravitational force distribution, and the same argument as for gravity applies: the buoyancy force does not cause rotation and must thus balance its torques out about a single point. That point is a good representative for where the buoyancy force as a whole can be modelled to act in. And this happens to be the exact same point as the centre-of-gravity (or centre-of-mass). Since this is the centre-of-geometry in case of water bodies, you can find this point as you've shown solely via geometric considerations.
A: The buoyancy forces on an object can be replaced by a single force acting at a point called the centre of buoyancy. If the object has a uniform density then the centre of buoyancy coincides with the centre of mass of the submerged part of the object. But if the object does not have a uniform density then the centre of buoyancy may be in a different place.
A: Both gravity and Archimedes' law act on each mass element with the same magnitude and direction.
If a force acts in the same way at all points in a body (e.g gravity in the approximation in which $\vec{F_g}\approx mg\hat{t}$ where $\hat{t}$ is a direction, usually the $z$-axis) then its effect is the same of that of a force acting on the center of mass of the body. This is not true for all forces, not even for gravity if you use the more general definition of $\vec{F_g}\sim {\hat{r}\over r^2}$. It is true for parallel force fields. Gravity close to the Earth's surface or Archimedes' force are good examples.
We first prove it in general, and then discuss the case of Archimedes' law.
Parallel forces - total force
To prove that, let's divide our body in small volumes $dV$, each of them having a mass $\rho(\vec{r}) dV$ where $\rho(\vec{r})$ is the density at position $\vec{r}$. Of course then the total mass of the body is
$$M=\int\rho(\vec{r}) dV$$ (where the integral is carried out on the whole body - I will not mark it, but all integrals here are carried out in the volume of the body.).
On each mass there is a force $d\vec{F}=dm g\hat{t}$ (where $dm$ is the mass, $\hat{t}$ the direction of the parallel force field and $g$ a constant which I chose to be $g$ so it resembles the gravitational case. (The following, with some modifications, is similar to the case where the constant force is not mass-dependent, but as in this case both gravity and Archimedes' are, I will keep the mass dependence clear).
Then the total force acting on the body is given by
$$\vec{F}=\int d\vec{F}=\int dm g\hat{t}=\int g\hat{t}\rho(\vec{r}) dV$$
now, because $\hat{t}$ is constant (does not depend on where you are in the body we are considering)
$$\vec{F}=g\hat{t}\int \rho(\vec{r}) dV$$ and the final integral is just the total mass of the body $M=\int \rho(\vec{r}) dV$ so the total force is te same we expect for a body of total mass $M$ i.e.
$$\vec{F}=Mg\hat{t}$$
Parallel forces - total torque and and point of action
The question now is: where is this force acting? To do that, we compute the torque $\vec{\tau}$ due to the force with respect to an arbitrary point $\vec{r}_0$. On an individual mass $dm$ at position $\vec{r}$ the torque will be
$$d\vec{\tau} = d\vec{F} \times (\vec{r}-\vec{r}_0)$$
where $\times$ is the vector product,
so that the total torque is
$$\vec{\tau} = \int d\tau = \int d\vec{F} \times (\vec{r}-\vec{r}_0)$$
integrating on the whole body, and again using $d\vec{F}=dm g\hat{t}$
$$\vec{\tau} = \int  \rho(\vec{r})dV g\hat{t} \times (\vec{r}-\vec{r}_0)$$
which becomes
$$\vec{\tau} = g \left(\int  \rho(\vec{r})dV \hat{t}\times\vec{r}-\int\rho(\vec{r})dV \hat{t}\times\vec{r}_0\right)$$
and finally, using the fact that $\vec{r}_0$ and $\hat{t}$ are constants in the body
$$ \vec{\tau} = g \left(\hat{t}\times \int  \rho(\vec{r})dV \vec{r}-\hat{t}\times\vec{r}_0 \int\rho(\vec{r})dV \right)$$
the right integral is just the total mass again, the left one can be rewritten by multiplying/dividng times $M$ as
$$\vec{\tau} = g \left(\hat{t}\times {M\over M} \int  \rho(\vec{r})dV \vec{r}-\hat{t}\times\vec{r}_0 M \right)$$
and the left integral now is the definition of center of mass  $\vec{r}_{cm}={1\over M}\int  \rho(\vec{r})dV \vec{r}$ so that
$$ \vec{\tau}= g \left(M \hat{t}\times \vec{r}_{cm}-M\hat{t}\times\vec{r}_0  \right)$$
and putting stuff together
$$\vec{\tau} = Mg\hat{t}\times(\vec{r}_{cm}-\vec{r}_{0})$$
so also the  total torque can be thought as of a total force $Mg\vec{t}$ acting on the center of mass (because the sum of all infinitesimal torques is equivalent to that of a single force acting on the center of mass as the individual positions $\vec{r}$ are substituted by a single position $\vec{r}_{cm}$).
Gravity vs Archimedes' law
Summing up, a mass-dependant parallel force field is equivalent to a force whose magnitude is that of the same force acting on a point-body of mass $M$ and whose position is on the center of mass.
For gravity (in the constant approximation)  that is all.
In the case of Archimedes law however, what counts is not the center of mass of the system but the center of mass of the part of the body which is submerged which, for bodies only partially submerged, does not coincide with the center of mass in total. This is because the integral above are limited to the part of body on which Archimedes' forces acts i.e. the parts below water.
This is why some bodies partially submersed in water (like the pole in your picture) are not stable and there will be a torque acting on them as gravity and Archimedes' force are not acting on the same point.
For a rod of homogeneous density, the center of mass is at ${L/over 2}$ but the center of mass of the part under water is at the half-length of the fraction below water (in your case half the distance between $A$ and $D$ i.e. $L_A=( {L\over 2\sin(\theta)})/2$.
A: You have correctly identified the centre of buoyancy which is the centre of gravity for the volume of liquid which the rod has displaced.
