Answer
$$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$
$K_c = 5.6 \times 10^{23}$
Work Step by Step
1. Multiply the first reaction by 2, to do so, raise $K_c^{'}$ to the power of 2.
$$2S(s) + 2O_2(g) \leftrightharpoons 2SO_2(g)$$
$K_c^{'} = (4.2 \times 10^{52})^2 = 1.764\times 10^{105}$
2. Invert the first reaction:
$$2SO_2(g) \leftrightharpoons 2S(s) + 2O_2(g)$$
$K_c^{'} = 1/(1.764 \times 10^{105}) = 5.7 \times 10^{-106}$
3. Add both reactions; removing the repeated components, the Kc of the resulting reaction is the multiplication of $K_c^{''}$ and the new $K_c^{'}$
$$2SO_2(g) +O_2(g) \leftrightharpoons 2SO_3(g)$$
$K_c = 5.7 \times 10^{-106} \times 9.8 \times 10^{128} = 5.6 \times 10^{23}$
4. Write the equilibrium constant expression:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$
