1. Write the given reactions and equilibrium constants

$\text{(1)}\quad S(s) + O_2(g) \rightleftharpoons SO_2(g), \qquad K_1 = 10^{52}$

$\text{(2)}\quad 2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g), \qquad K_2 = 10^{129}$

We need the equilibrium constant for:

$\text{(3)}\quad 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

2. Manipulate the given equations to obtain the target reaction

First, multiply reaction (1) by 2:

$2S(s) + 2O_2(g) \rightleftharpoons 2SO_2(g)$

When a reaction is multiplied by 2, its equilibrium constant becomes:

$K = (K_1)^2 = (10^{52})^2 = 10^{104}$

Now reverse this reaction:

$2SO_2(g) \rightleftharpoons 2S(s) + 2O_2(g)$

On reversing, the equilibrium constant becomes:

$K = \frac{1}{10^{104}} = 10^{-104}$

3. Add this reversed reaction to reaction (2)

Reaction (2):

$2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g), \qquad K_2 = 10^{129}$

Adding:

$2SO_2(g) \rightleftharpoons 2S(s) + 2O_2(g)$

and

$2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g)$

Cancel $2S(s)$ from both sides:

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

which is exactly the required reaction.

4. Calculate the equilibrium constant

For added reactions, equilibrium constants multiply:

$K = K_2 \times \frac{1}{(K_1)^2}$

Substitute the values:

$K = 10^{129} \times 10^{-104} = 10^{25}$

5. Match with the options

$K = 10^{25}$

So the correct option is:

B: $10^{25}$

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

Hence, the derived answer agrees with the stored correct answer.