How do I calculate the equilibrium constant of a reaction from two related reactions' constants?
Equilibrium constants depend upon the choice of reaction equations. I have noted the extended way to calculate $K''_\mathrm{c}$ below:

$$\begin{align}\ce{2S + 3 O2 &amp;&lt;=&gt; 2 SO3}\tag{2}\\[0.4em]
K''_\mathrm{c} &amp;= \frac{[\ce{SO3}]^2}{[\ce{O2}]^3}\end{align}$$

(Sulfur does not take part in the equilibrium constant because its activity, as a solid, is $a(\ce{S}) = 1$.)

Having this, we can approach the question. First, we need to add particles in an equivalent way (i.e. the same to reactant and product side) to get from equation $(3)$ to something that can be expressed as a linear combination of $(1)$ and $(2)$ ($a(1) + b(2)$):

\begin{align}\ce{2SO2 + O2 &amp;&lt;=&gt; 2SO3}\tag{3}\\[0.4em]
\ce{2SO2 + 3 O2 &amp;&lt;=&gt; 2 SO3 + 2 O2}\tag{$3'$}\\[0.4em]
\ce{2 SO2 + 3 O2 + 2 S &amp;&lt;=&gt; 2 SO3 + 2 O2 + 2 S}\tag{$3''$}\end{align}


   $\displaystyle (3'') = -2 \times (1) + 1 \times (2)$


With that in hand, we can determine what the equilibrium constant of $(3'')$ would look like and how that can be traced back to the other equilibrium constants:


   $$\displaystyle K_\mathrm{c} = \frac{[\ce{SO3}]^2[\ce{O2}]^2}{[\ce{SO2}]^2[\ce{O2}]^3}\tag{4}$$

 $$\displaystyle K_\mathrm{c} = \frac{[\ce{O2}]^2}{[\ce{SO2}]^2} \times \frac{[\ce{SO3}]^2}{[\ce{O2}]^3} = (K'_\mathrm{c})^{-2} \times K''_\mathrm{c} \tag{5}$$


And thus, your final solution is:


   $$(4.2\times10^{52})^{-2} \times (9.8\times10^{128}) = 5.6\times10^{23}\tag{6}$$