At a certain temperature, 13.0 moles of SO3 are placed into a rigid 2.0 L container.
The SO3 dissociates as follows:
2 SO3 (g) ⇌ 2 SO2 (g) + O2 (g)

At equilibrium 10 moles of SO3 are present.
Calculate Keq at this temperature.

I got that K=[SO2]^2*[O2]^1/[SO3]^2
but that's all I have

Yes, that's enough to work out the equilibrium constant of the reaction.

I'll not give you the answer directly, we'll work through it.

You had 13 moles of SO3 initially in a 2L container.

At equilibrium, there are 10 moles of SO3 left. What is the concentration of the SO3?
How many moles of SO2 were produced?
How many moles of O2 were produced?
What are their respective concentrations?

So, I think 3 moles of SO3 was used, so the concentration of SO3 should be 3moles/2L=1.5M, and 3 moles of SO2 were produced, and 1.5 moles of O2 were produced. And the concentration for SO2 and O2 should be 1.5 M, and 0.75M, respectively, am I right?

Yes for SO2 and O2.

But there are 10 moles of SO3 which remain, not 3 moles ;)

Now that you got it, use your formula to find the equilibrium constant.

But isn't is remaining means not reacting? Then why we are using 10 moles for SO3 for one part but using 3 moles of SO3 to get moles of SO2 and O2 for the other part?

What you need to find the equilibrium constant is the amount which remain in each case.

There is dynamic equilibrium in the system, so that the reactants not really remain unreacted.

On the right hand side, you you have 3 moles of SO2 which remain and 1.5 mole of O2 which remain too. Remember that it is a reversible reaction and that SO2 and O2 are reacting too to form SO3.

Since you took what remained of SO2 and O2, you take what remains of SO3 also!