If $\bar{a}, \overline{\mathrm{~b}}, \overline{\mathrm{c}}$ are three vectors such that $|\bar{a}|=\sqrt{31}, 4|\overline{\mathrm{~b}}|=|\overline{\mathrm{c}}|=2$ and $2(\bar{a} \times \bar{b})=3\left(\bar{c} \times \bar{a}\right.$ )and if the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2 \pi}{3}$ then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$

Given vector magnitudes:

$|\bar{a}| = \sqrt{31}$

$|\bar{b}| = \frac{1}{2}$

$|\bar{c}| = 2$

The angle between $\bar{b}$ and $\bar{c}$ is $\theta = \frac{2\pi}{3}$.

The vector relation $2(\bar{a} \times \bar{b}) = 3(\bar{c} \times \bar{a})$ becomes $2(\bar{a} \times \bar{b}) + 3(\bar{a} \times \bar{c}) = \overline{0}$ after using $\bar{c} \times \bar{a} = -(\bar{a} \times \bar{c})$.

Factoring gives $\bar{a} \times (2\bar{b} + 3\bar{c}) = \overline{0}$, implying $2\bar{b} + 3\bar{c} = k\bar{a}$ for some scalar $k$.

The magnitude squared yields:

$|2\bar{b} + 3\bar{c}|^2 = 4|\bar{b}|^2 + 9|\bar{c}|^2 + 12(\bar{b} \cdot \bar{c}) = 4(\frac{1}{4}) + 9(4) + 12(-\frac{1}{2}) = 1 + 36 - 6 = 31$

Since $|k\bar{a}|^2 = k^2|\bar{a}|^2 = k^2 \cdot 31$, we find $k^2 = 1$, so $k = \pm 1$.

For both cases, $|\bar{a} \times \bar{c}|^2 = 4|\bar{b} \times \bar{c}|^2 = 4(|\bar{b}||\bar{c}|\sin\theta)^2 = 4(\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{3}}{2})^2 = 3$.

When $k = 1$, $\bar{a} \cdot \bar{b} = 2|\bar{b}|^2 + 3(\bar{b} \cdot \bar{c}) = 2(\frac{1}{4}) + 3(-\frac{1}{2}) = -1$.

When $k = -1$, $\bar{a} \cdot \bar{b} = -[2|\bar{b}|^2 + 3(\bar{b} \cdot \bar{c})] = 1$.

The expression simplifies to:

$\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2 = \frac{|\bar{a} \times \bar{c}|^2}{(\bar{a} \cdot \bar{b})^2} = \frac{3}{1} = 3$

Final answer: $\boxed{3}$