$2S{O}_2\left(g\right) + O_2\left(g\right) \to 2S{O}_3\left(g\right).$   In an equilibrium mixture, the partial pressures are $P_{S{O}_3} = 43Kpa ; P_{O_2} = 530K pa and P_{S{O}_2} = 45Kpa$The equilibrium constant Kp = × ${10}^{-2}$.

$$\begin{gathered} 2S{O}_2\left(g\right) + O_2\left(g\right) \to 2S{O}_3\left(g\right). \\ {K}_{p }=\frac{{\left(pS{O}_{3\left(g\right)}\right)}^2}{pS{O}_{2\left(g\right)}}\times p{O}_{2\left(g\right)} \\ = \frac{43\times 43}{45\times 45}\times 530 P{a}^{-1} \\ = 172.28\times {10}^{-5} P{a}^{-1} \\ = 17228\times {10}^{-2 } atm \end{gathered}$$

Hence, the correct answer is $17228\times {10}^{-2 } atm$