At 330 K and a total pressure of 1.00 atm, an equilibrium mixture of N2O4(g)rightleftharpoons 2NO2(g) has an average molar mass of 69.0 g mol^-1…

For one mole of $\mathrm{N_2O_4}$ dissociating to a degree $\alpha$, the moles at equilibrium are $(1-\alpha)+2\alpha=1+\alpha$, so the average molar mass is $M=92/(1+\alpha)$. Setting $92/(1+\alpha)=69.0$ gives $1+\alpha=1.333$ and $\alpha=1/3$. The equilibrium constant is $K_p=\dfrac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}}=\dfrac{4\alpha^2P}{1-\alpha^2}$, which for $P=1.00$ atm and $\alpha=1/3$ is $K_p=\dfrac{4/9}{8/9}=0.500$ atm. Since $K_p$ depends only on temperature, for $\alpha=0.200$ we need $P=\dfrac{K_p(1-\alpha^2)}{4\alpha^2}=\dfrac{0.500\times0.96}{0.16}=3.00$ atm. As expected from Le Chatelier's principle, a higher pressure suppresses dissociation, from $\alpha=0.333$ to $0.200$. The route is therefore: use the molar mass to find the degree of dissociation, convert it to $K_p$ at the given pressure, and then reuse $K_p$ at the new pressure, because the equilibrium constant is fixed at constant temperature. The answer is $3$.