The standard equilibrium constant for the reaction ${\mathrm{SO}}_3( \mathrm{g})\rightleftharpoons {\mathrm{SO}}_2( \mathrm{g})+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})$ is $K_{\mathrm{c}}°=4.9\times {10}^{-2}$ the value of $K_{\mathrm{c}}°$ the reaction $2{\mathrm{SO}}_2( \mathrm{g})+{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons 2{\mathrm{SO}}_3( \mathrm{g})$ will be

The reaction $2{\mathrm{SO}}_2( \mathrm{g})+{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons 2{\mathrm{SO}}_3( \mathrm{g})$ is obtained by reversing the reaction ${\mathrm{SO}}_3( \mathrm{g})\rightleftharpoons {\mathrm{SO}}_3( \mathrm{g})+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})$ followed by multiplication by 2. Hence

$K_c°=\frac{1}{{\left(K_c°\right)}^2}=\frac{1}{{\left(4.9\times {10}^{-2}\right)}^2}=416$