If a 3.30 m sample of a is heated to 500 k, what is the concentration of b at equilibrium?

(Missing in your question):

we have the following reaction:

A (aq) ↔ 2 B (aq)

and Kc = 7.02 x 10^-6 at 500K

So at equilibrium,

Kc = [Products] / [ reactants]

= [B]^2 / [A]

we have [A] = 3.3 m and Kc is given = 7.02 x10^-6

by substitution:

7.02x10^-6 = [B]^2 / 3.3

∴[B]^2 = 2.3 x 10^-5

∴[B] = 0.005 m