The problem asks to analyze the properties of two spherical conductors, a shell A and a solid sphere B, after they are connected by a thin metal wire. Let the final charges on them be $Q_A$ and $Q_B$, and their radii be $R_A$ and $R_B$ respectively, with the given condition $R_A > R_B$. The initial charge on each was $+Q$, so the total charge of the system is $2Q$, which is conserved. Thus, $Q_A + Q_B = 2Q$.

When the two conductors are connected by a wire, charge flows between them until they reach the same electrostatic potential. Let this common potential be $V$.

For a spherical conductor of radius $R$ and charge $q$, the potential on its surface is given by $V = \frac{1}{4\pi\epsilon_0}\frac{q}{R}$. Since the potentials of A and B are equal: $V_A = V_B$ $\frac{1}{4\pi\epsilon_0}\frac{Q_A}{R_A} = \frac{1}{4\pi\epsilon_0}\frac{Q_B}{R_B}$ $\frac{Q_A}{R_A} = \frac{Q_B}{R_B} \implies \frac{Q_A}{Q_B} = \frac{R_A}{R_B}$

Now, let's evaluate each option based on this result.

Option A: $E_A^{inside} = 0$

1. Sphere A is a metallic (conducting) shell.
2. A fundamental property of conductors in electrostatic equilibrium is that the electric field inside the conductor is zero. This applies to both the material of the shell and the hollow region (cavity) inside it, as there are no charges placed within the cavity.
3. Therefore, $E_A^{inside} = 0$. This statement is correct.

Option B: ${Q_A} > {Q_B}$

1. From our potential equalization condition, we derived the relationship between the final charges: $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$
2. The problem states that $R_A > R_B$.
3. This implies that the ratio $\frac{R_A}{R_B} > 1$.
4. Therefore, $\frac{Q_A}{Q_B} > 1$, which means $Q_A > Q_B$. The larger sphere holds more charge.
5. This statement is correct.

Option C: ${{{\sigma _A}} \over {{\sigma _B}}} = {{{R_B}} \over {{R_A}}}$

1. The surface charge density ($\sigma$) is defined as charge per unit area. For a sphere of radius $R$, the surface area is $4\pi R^2$.
2. For sphere A: $\sigma_A = \frac{Q_A}{4\pi R_A^2}$
3. For sphere B: $\sigma_B = \frac{Q_B}{4\pi R_B^2}$
4. Let's find the ratio of the surface charge densities: $\frac{\sigma_A}{\sigma_B} = \frac{Q_A / (4\pi R_A^2)}{Q_B / (4\pi R_B^2)} = \frac{Q_A}{Q_B} \cdot \frac{R_B^2}{R_A^2}$
5. We already know that $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$. Substituting this into the equation: $\frac{\sigma_A}{\sigma_B} = \left(\frac{R_A}{R_B}\right) \cdot \left(\frac{R_B^2}{R_A^2}\right) = \frac{R_B}{R_A}$
6. This statement is correct.

Option D: $E_A^{on\,surface} < \,E_B^{on\,surface}$

1. The electric field at the surface of a charged conducting sphere is given by $E = \frac{\sigma}{\epsilon_0}$.
2. So, $E_A = \frac{\sigma_A}{\epsilon_0}$ and $E_B = \frac{\sigma_B}{\epsilon_0}$.
3. The ratio of the electric fields is: $\frac{E_A}{E_B} = \frac{\sigma_A / \epsilon_0}{\sigma_B / \epsilon_0} = \frac{\sigma_A}{\sigma_B}$
4. From our analysis of Option C, we know $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A}$.
5. Therefore, $\frac{E_A}{E_B} = \frac{R_B}{R_A}$.
6. Since it is given that $R_B < R_A$, the ratio $\frac{R_B}{R_A} < 1$.
7. This means $\frac{E_A}{E_B} < 1$, which implies $E_A^{on\,surface} < E_B^{on\,surface}$. The electric field is stronger on the surface of the smaller sphere.
8. Alternatively, we know the common potential is $V$. The electric field on the surface can also be expressed as $E = \frac{V}{R}$. Thus, $E_A = \frac{V}{R_A}$ and $E_B = \frac{V}{R_B}$. Since $R_A > R_B$, it follows that $E_A < E_B$.
9. This statement is correct.

Since all four options A, B, C, and D are correct, this is a multiple correct question where all options are valid.