There is an important thing to note about your question, which is the understanding the relationship between math and physics. Physics uses mathematical equations to describe reality. Physics equations are approximations and may assume certain conditions (pressure, temperature, etc.) to offer an equation that is strictly true under those certain conditions. Thus, physics equations like the one you provided has boundaries in where its true.
In order to find the average velocity for real gases, you'll need to strictly define the conditions your gas is under. Although the average velocity of gases given by Maxwell-Boltzman distribution may be accurate, it may still deviate from a real gas you are asking about because it may be under different conditions. You may need to add or subtract some term to the equation you provided in order for it to match your real gas.
In short, there is no corresponding velocity for a real gas in the sense that "no physics equation to date is perfectly accurate in describing reality."
You may have a very accurate and precise equation that corresponds to reality, but it still serves as an approximation. What's great is that depending upon the system you are studying, sometimes you don't need such a large accuracy and your physics equation may have an accuracy that may exceed the system in question. This would be the case where your physics equation would be equivalent to describing reality, but you really need to specify your system first.
Moreover, since the equation is describing an average velocity, it means that you could have a distribution of velocities that has a large spread or a small spread.
The main difference between gases and solids is indeed that in the latter, interactions between constituents are not negligible.
However, i do not think that your approach is valid, because the right result should be (using your notations) $K_s = K_g$.
Here is an approximate explanation :
The situation of a solid can be modelled this way : the interaction between two particles of this solid is described by a potential $U$ that is a function of their relative position. For reasonably low temperatures, the particles will remain close to their equilibrium position and oscillate thermally around it.
For small deviations around this position, the position can be approximated to be $$U(\vec{x}) \simeq \frac{1}{2} k \vec{x}^2$$
Where $x$ is the deviation of a particle from its equilibrium position.
For oscillators, the energy is equally distributed between potential and kinetic energy so that :
$$\frac{1}{2}m\langle \vec{v}^2 \rangle = \frac{1}{2}k \langle \vec{x}^2\rangle$$
As stated by the equipartition theorem, to each degree of freedom of the system corresponds an average energy $\frac{1}{2} k_B T$. So, just like in gases, the average kinetic energy is still given by :
$$\frac{1}{2}m\langle \vec{v}^2 \rangle = \frac{3}{2} k_B T$$
What really differs is that when interactions aren't negligible anymore, the total energy is the sum of the kinetic energy and the potential energy which is on average (using the two previous equations) $$\frac{3}{2} k_B T + \frac{3}{2} k_B T = 3 k_B T$$
This might sound surprising that molecules can go as fast in a solid as in a gas, but note that in a solid they don't go further than $\langle \vec{x} \rangle$ on average and their high kinetic energy is due to their oscillation frequency $\omega = \sqrt{\frac{k}{m}}$.
Side note
Since the average thermal energy for one particle of the solid is $3k_B T$, the average energy for $N$ solids is $3Nk_B$. As a result the heat capacity of a solid in $J/K/mol$ should approximately be around $3R$, and this result, known as the Dulong–Petit Law, has some experimental support, giving some credit to this approach.
Best Answer
As stated by @Wolphram jonny my mistake was in using blindly the formula without questioning the nature of what I called $m$.