For the equilibrium in gaseous phase in 2 lit flask, we start with 2 moles of  $S{O}_2$   and 1 mole of   $O_2$ at 3 atm   $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)}.$  When  equilibrium   is attained, Pressure changes to 2.5 atm. Hence equilibrium constant  $K_C$   is

 $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)}$
2               1            0
$2-2x 1-x 2x$
Moles at equilibrium  $=3-x$
At equilibrium pressure becomes  2.5 atm 
Since pv = n RT
Hence      $p\alpha n$  
Thus  $\frac{3-x}{3}=\frac{2-5}{3},x=0.5$
$$\begin{gathered} \left[S{O}_2\right]=\frac{2-2x}{2}=0.5M \\ \left[O_2\right]=\frac{1-x}{2}=0.25M;\left(S{O}_3\right)=\frac{2x}{2}=0.5M \\ {K}_c=\frac{{\left[S{O}_3\right]}^2}{{\left[S{O}_2\right]}^2\left[O_2\right]}=\frac{{\left(0.5\right)}^2}{{\left(0.5\right)}^2\times 0.25}=4 \end{gathered}$$