How to find inertia tensor of a circular ring from angular momentum and velocity? 
Consider a thin circular ring with radius $R$ and axis of rotation as shown in the figure. If $\vec{L}$ denotes angular momentum and $\vec{w}$ is the angular velocity then
$$\vec{L}=\begin{bmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{xy} & I_{yy} & I_{yz}\\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix} \vec{w}$$
Where the matrix is also called as moment of inertia tensor.
MY QUESTION:-
Is it possible to find the moment of inertia of this ring along the given axis using the equation I mentioned?
 A: The perfectly aligned MMOI matrix for a ring that rotates about the y axis is
$$ \mathcal{I} = \begin{bmatrix}
 \frac{m}{2} \left( \frac{h^2}{2} + r^2 + \frac{w^2}{4} \right) & & \\ & \frac{m}{2} \left( 2 r^2 + \frac{w^2}{2} \right) & \\ & & \frac{m}{2} \left( \frac{h^2}{2} + r^2 + \frac{w^2}{4} \right) \end{bmatrix}$$
where $r$ is the radius of the section centroid, $w$ is the width along the x axis, and $h$ is the height along the y axis.
As you can see the top left element equals the bottom right element.
Now if angular momentum is known, even though the MMOI matrix is diagonal, there still isn't enough information to extract the geometry from below
$$ \begin{bmatrix} L_x \\ L_y \\ L_z \end{bmatrix} = \mathcal{I} \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix} $$
This is because the ratio of momentum along the x axis and z axis is
$$ \frac{L_x}{L_z} = \frac{\omega_x}{\omega_z}$$
and all the geometry terms cancel out.
But if we know the width $w$, then we can solve the momentum equations for
$$ \begin{aligned}
 r & = \sqrt{ \frac{L_y}{m \omega_y} - \frac{w^2}{4} } \\
 h & = \sqrt{ \frac{4 L_x}{m \omega_x} - \frac{2 L_y}{m \omega_y} } \end{aligned} $$
A: 
Now is it possible to find the moment of inertia of this ring along the given axis using the equation I mentioned?

In general, you cannot determine inertia tensor from angular moment and velocity vectors. There are 6 unknowns and only 3 linear equations. You can rewrite the equation for angular momentum
$$
\left[
\begin{array}{cccccc}
\omega_x & 0 & 0 & \omega_y & 0 & \omega_z \\
0 & \omega_y & 0 & \omega_x & \omega_z & 0 \\
0 & 0 & \omega_z & 0 & \omega_y & \omega_x \\
\end{array}
\right]
\left[
\begin{array}{c}
I_{xx} \\
I_{yy} \\
I_{zz} \\
I_{xy} \\
I_{yz} \\
I_{zx}
\end{array}
\right]
=
\left[
\begin{array}{c}
L_x \\ L_y \\ L_z
\end{array}
\right] \tag 1
$$
What does linear algebra say - does the above equation have unique solution in vector $I$? The best you can do here is to express 3 moments of inertia as a function of other 3 moments of inertia. What this means is that if any 3 moments of inertia are known (at least), you can determine remaining unknown moments of inertia.

Object is symmetric about all axes
If the object is symmetric about all axes, then all products of inertia are zero ($I_{xy}$, $I_{yz}$, and $I_{zx}$). With this the Eq. (1) becomes
$$
\left[
\begin{array}{cccccc}
\omega_x & 0 & 0 \\
0 & \omega_y & 0 \\
0 & 0 & \omega_z \\
\end{array}
\right]
\left[
\begin{array}{c}
I_{xx} \\
I_{yy} \\
I_{zz}
\end{array}
\right]
=
\left[
\begin{array}{c}
L_x \\ L_y \\ L_z
\end{array}
\right] \tag 2
$$
and the moments of inertia are
$$
\left[
\begin{array}{c}
I_{xx} \\
I_{yy} \\
I_{zz}
\end{array}
\right]
=
\left[
\begin{array}{cccccc}
\omega_x & 0 & 0 \\
0 & \omega_y & 0 \\
0 & 0 & \omega_z \\
\end{array}
\right]^{-1}
\left[
\begin{array}{c}
L_x \\ L_y \\ L_z
\end{array}
\right]
=
\left[
\begin{array}{c}
L_x/\omega_x \\ L_y/\omega_y \\ L_z/\omega_z
\end{array}
\right]
 \tag 3
$$

Moment of inertia about axis through the circle center
Here I show how to find moment of inertia of an uniform circular ring about axis that goes through the circle center and perpendicular to the ring surface
$$I = \iiint r^2 dm = \iiint r^2 \rho dV = \int_{z=-t/2}^{z=t/2} \int_{\phi=0}^{\phi=2\pi} \int_{r-\Delta r}^{r} r^2 \rho r dr d\phi dz = 2\pi t \rho \int_{r-\Delta r}^{r} r^3 dr$$
where $t$ is thickness of the ring. The above integral results in
$$I = 2\pi t \rho \frac{1}{4} \Bigl. r^4 \Bigr|_{r-\Delta r}^{r} = \frac{1}{2} \underbrace{\bigl( r^2 - (r-\Delta r)^2 \bigr) \pi t \rho}_{M} \bigl( r^2 + (r-\Delta r)^2 \bigr) = $$
$$= \frac{1}{2} M \bigl( 2r^2 - 2r\Delta r + (\Delta r)^2 \bigr) = M r^2 \Bigl(1 - \bigl(\frac{\Delta r}{r}\bigr) + \frac{1}{2} \bigl(\frac{\Delta r}{r}\bigr)^2 \Bigr)$$
where $M$ is mass of the ring. For $r \gg \Delta r$ the above expression is simplified into $I \approx Mr^2$.
