I confirm your answer.

The molar density of the mixture is $$\frac{n}{V}=\frac{p}{RT}=0.0326\ moles/liter$$So, the average molecular weight of the mixture is $$\frac{2}{0.0326}=61.35\ grams/mole$$If x is the mole fraction of NO2 and (1-x) is the mole fraction of N2O4, then the average molecular weight of the mixture is also $$46x+92(1-x)=61.35$$. Solving for x gives x = 1/3. So, (1-x)=2/3. These are also the partial pressures of NO2 and N2O4, respectively (in atm). This leads to the values of Kp and Kc that you calculated.

I confirm your answer.

The molar density of the mixture is

$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$

So, the average molecular weight of the mixture is

$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$

If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also

$$46x + 92 (1 - x) = 61.35$$

Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.