To solve this problem, we need to use the equilibrium constant expression for the reaction:$\\ 3O_2(g) \rightleftharpoons 2O_3(g) \\$The equilibrium constant $K_c$ is given by:$\\ K_c = \frac{[O_3]^2}{[O_2]^3} \\$Given:$\\ K_c = 3.0 \times 10^{-59} \\ [O_2] = 0.040 \,$M$\\$We need to find the concentration of $O_3$ at equilibrium, denoted as $[O_3]. \\$Substitute the known values into the equilibrium expression:$\\ 3.0 \times 10^{-59} = \frac{[O_3]^2}{(0.040)^3} \\$Solve for $[O_3]^2: \\ [O_3]^2 = 3.0 \times 10^{-59} \times (0.040)^3 \\$Calculate $(0.040)^3: \\ (0.040)^3 = 0.040 \times 0.040 \times 0.040 = 6.4 \times 10^{-5} \\$Substitute back to find $[O_3]^2: \\ [O_3]^2 = 3.0 \times 10^{-59} \times 6.4 \times 10^{-5} \\ [O_3]^2 = 1.92 \times 10^{-63} \\$Take the square root to find $[O_3]: \\ [O_3] = \sqrt{1.92 \times 10^{-63}} \\ [O_3] = 4.38 \times 10^{-32} \,$M$\\$Therefore, the concentration of $O_3$ at equilibrium is $4.38 \times 10^{-32} \,$M$. \\$This corresponds to Option 1.