At every point along the beam, the curvature has to be such that the externally applied bending moment exactly counters the internal stress. This tells you that the curvature is not constant - it is a function of distance to the side (largest in the middle, zero at the wall). This means that your assumption of "circular section" is wrong.
See for example figure 3.16 in this link and associated derivations.
Simplifying the description found there:
From their equation 3.21, the curvature $\rho$ of a beam is related to the bending moment $M$ by
$$\rho = \frac{EI}{M}\tag1$$
Where $E$ is the Young's modulus and $I$ is the second moment of area. For a rectangular beam (not specified in your question, but that's what I am assuming) we can compute $I$ as
$$I = \frac{bd^3}{12}\tag2$$
(see for example this link)
Now we need an expression for the bending moment as a function of position. For points to the left of the center, bending moment is proportional to $Wx/2$ - half the weight (two supports) times the distance from the support.
Knowing that the radius of curvature is (for small deflections) inversely proportional to the second derivative of the shape, we can write
$$\frac{d^2 y}{dx^2} = -\frac{Wx}{2EI}\tag3$$
Integrating twice, we get
$$y = -\frac{Wx^3}{12EI} + Ax + B\tag4$$
If we set $y=0$ at $x=0$, we get $B=0$. Setting the slope of the curve =0 at $x = \frac{\ell}{2}$, we find
$$-\frac{W(\ell/2)^2}{4EI} + A = 0$$
$$A = \frac{W\ell^2}{16EI} \tag5$$
leading to an expression for the deflection
$$y = \frac{Wx}{12EI}\left(\frac{3\ell^2}{4}-x^2\right) \tag6$$
Substituting $x=\frac{\ell}{2}$ into (6), and using expression (2) for $I$, we obtain the deflection you were looking for.
This expression agrees with equation (7) at this reference.
Strictly speaking, Young's modulus is not always greater than the shear modulus, but it does tend to work out that way. You can see the reason why if you look at the relation between the two quantities (and Poisson's ratio).
$$
G = \frac{E}{2(1+\nu)}
$$
Combined with the knowledge that $\nu$ can be anywhere in the range $(-1, \frac{1}{2})$, one can see that G can be greater than E for $\nu < -1/2$. That being said, materials with such a negative Poisson's ratio are extremely uncommon, and it is safe to assume that the shear modulus is less than half of Young's modulus.
For an explanation of why the Poisson's ratio must fall within the above range, I invite you to check out some of my previous answers.
High-level: Range of poissons ratio
Detailed explanation: Limits of Poisson's ratio in isotropic solid
Best Answer
For simplicity let's consider a cubic shape of side $\ell$, so the volume is $\ell^3$.
Suppose we compress the cube so its volume decreases from $V$ to $V - \delta V$, and suppose this needs a pressure $P$.
Then the bulk modulus is:
$$ K = -\delta P \frac{V}{\delta V} $$
in the limit of $\delta V \rightarrow 0$. Now suppose we deform the cube slightly by pushing the top of it to one side. Suppose this takes a force $F$:
Then the shear modulus is:
$$ G = \frac{F}{A} \frac{\ell}{x} $$
where $A$ is the area of the top face of the cube.
The first process involves compression only and no shear, while the second process involves shear only and no compression, so there is no reason why the bulk and shear moduli should necessarily be related. Of course in practice they are related because both originate from the interatomic forces in the material.
You ask:
but this is not a well defined question. Remember that compression and shear are unrelated processes. You could ask:
and this we can answer because if $G < K$ then the answer is yes, provide we equate the pressure in compression with the force per unit area in shear. However even though both have the units of pressure I'm not sure I would really say they were equivalent.