The given balanced chemical equation is: $3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g)$

For this reaction, the equilibrium constant expression ($K_c$) is written as the ratio of the concentration of the product raised to its stoichiometric coefficient to the concentration of the reactant raised to its stoichiometric coefficient: $K_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}$

Given the values at equilibrium: $K_c = 3.0 \times 10^{-59}$ $[\text{O}_2] = 0.040\text{ M} = 4 \times 10^{-2}\text{ M}$

Substituting these values into the equilibrium constant expression: $3.0 \times 10^{-59} = \frac{[\text{O}_3]^2}{(4 \times 10^{-2})^3}$ $[\text{O}_3]^2 = 3.0 \times 10^{-59} \times (4 \times 10^{-2})^3$ $[\text{O}_3]^2 = 3.0 \times 10^{-59} \times 64 \times 10^{-6}$ $[\text{O}_3]^2 = 192 \times 10^{-65}$ $[\text{O}_3]^2 = 19.2 \times 10^{-64}$

Taking the square root on both sides to find the concentration of ozone: $[\text{O}_3] = \sqrt{19.2 \times 10^{-64}} = \sqrt{19.2} \times 10^{-32}$ Since $\sqrt{16} = 4$ and $\sqrt{25} = 5$, $\sqrt{19.2}$ is approximately $4.38$. $[\text{O}_3] \approx 4.38 \times 10^{-32}\text{ M}$

Thus, the concentration of $\text{O}_3$ is $4.38 \times 10^{-32}\text{ M}$.