Calculate pressure equilibrium constant from concentration equilibrium constant
If you intend to get an answer in Pascal, which is an SI unit, you need to have your original values in SI units as well.

Note that:

$$K_\mathrm c=\pu{4.2M^-2}=4.2~\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}=4.2~\left(\frac{\text{moles}}{\pu{10^-3m^3}}\right)^{-2}$$

The conversion factor for litre (non-SI) to metre cubed (SI unit) is $10^{-3}$. Since you have $K_\mathrm c$ in units of $\pu{M^-2}$, the final answer would be $10^{-6}$ of the answer you had arrived at.



Elaboration:

Let's do dimensional analysis on the RHS factor-by-factor. 


$K_\mathrm c$ has the units $\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}$ as shown above.
$RT$ has the units = $\pu{Jmol^-1T^-1}\cdot\pu{T}=\pu{J\cdot mol^-1}$
$(RT)^{-2}$ has the units = $\pu{J^-2\cdot mol^2}$.  
Since one joule $=\pu{Pa\cdot m^3}$, hence $(RT)^{-2}$ has the units = $\pu{Pa^-2\cdot m^-6\cdot mol^2}$
Multiply $K_\mathrm c$ and $(RT)^{-2}$ we get 
$$\text{Units of }K_\mathrm p=\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}\cdot\pu{Pa^-2\cdot m^-6\cdot mol^2}=\pu{Pa^-2\cdot m^-6\cdot litre^2}$$


Thus your final answer is not in the units of $\pu{Pa^-2}$ but rather $\pu{Pa^-2\cdot m^-6\cdot litre^2}$. Can you now see why we need the conversion factor?