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.