Here's how you'd go about solving this problem.

Start with the ICE table for the equilibrium reaction

#" "N_2O_(4(g)) rightleftharpoons 2NO_(2(g))#

**I**..........#n_0#..................0

**C**.......#-epsilon#..............#+2epsilon#

**E**...#n_0-epsilon#..............#2epsilon#

Express the mole fraction of the two species like this

#chi_(N_2O_4) = (n_0 - epsilon)/(n_0 - epsilon + 2epsilon) = (n_0 - epsilon)/(n_0 + epsilon)#

#chi_(NO_2) = (2epsilon)/(n_0 + epsilon)#

Now just use the definition of #K_p# to solve for #epsilon#

#K_p = P_(NO_2)^(2)/P_(N_2O_4) = ((2epsilon)^(2)/(n_0 + epsilon)^(2) * P^(2))/((n_0 - epsilon)/(n_0 + epsilon) * P)#

#K_p = ((2epsilon)^(2) * P^(cancel(2)))/((n_0 + epsilon)^(cancel(2))) * cancel(n_o + epsilon)/((n_0 - epsilon) * cancel(P))#

#K_p = (4epsilon^(2) * P)/((n_o - epsilon)(n_o + epsilon)) = (4epsilon^(2) * P)/(n_0^(2) - epsilon^(2))#

#K_p(n_0^(2) - epsilon^(2)) = 4epsilon^(2) P#

#4epsilon^(2)P + epsilon^(2)K_p - n_0^(2)K_p = 0#

#epsilon^(2)(4P + K_p) = n_0^(2) K_p | : n_0^(2)#

#epsilon^(2)/n_0^(2) = K_p/((4P + K_p)) |sqrt#

#color(green)(epsilon/n_0 = (K_p/(4P + K_p))^(1/2))#