Here's how you'd figure this out. First of all, your first equilibrium reaction is actually
2SO_(3(g)) rightleftharpoons 2SO_(2(g)) + O_(2(g)) color(blue)((1))
Now take a look at the second equilibrium reaction.
SO_(2(g)) + 1/2O_(2(g)) rightleftharpoons SO_(3(g)) color(blue)((2))
Notice that this time, sulfur trioxide, SO_3 is on the reactants' side, as opposed to the first equilibrium reaction in which it was on the products' side.
Moreover, the stoichiometric coefficients are halved for all the species that take part in the reaction, i.e. you have 1 mole of SO_3 instead of 2 moles, 1 mole of SO_2 instead of 2 moles, and 1/2 moles of O_2 instead of 1 mole.
So, start by writing the equilibrium constant for the second equilibrium, K_(c2)
K_(c2) = ([SO_3])/([SO_2] * [O_2]^"1/2")
The equilibrium constant for the first reaction, K_(c1), will be
K_(c1) = ([SO_2]^2 * [O_2])/([SO_3]^2)
You need to figure out how to write K_(c1) using K_(c2). The first thing you need to do is get sulfur trioxide in the denominator, since the first equilibrium features SO_3 as a reactant, not as a product. Flip K_(c2) to get
1/K_(c2) = ([SO_2] * [O_2]^"1/2")/([SO_3])
Now you need to get the exponents to match those of K_(c1). Notice that if you raise 1/K_(c2) to the power of 2, you get
1/(K_(c2))^2 = ([SO_2]^2 * [O_2]^(1/2 * 2))/([SO_3]^2) = ([SO_2] * [O_2])/([SO_3]^2)
But you know that
([SO_2]^2 * [O_2])/([SO_3]^2) = K_(c1)
Therefore,
1/K_(c2)^2 = K_(c1) => K_(c1) = color(green)(1/49^2)
