At equilibrium, 0.170 mol of o2 is present. calculate kc.

(missing in your question):
0.66 mol of SO3 is placed in 4.5 L container according to this reaction 
2SO3(g) ↔ 2SO2(g) + O2(g)
So the
for the equilibrium reaction so:
Kc = [SO2]^2 * [O2] / [ SO3]^2
when concentration (c) = n / V when n= no.of moles / v (volume per L)
∴ [ SO3]° = 0.66 / 4.5 = 0.147 M
by using Ice table we can determine the concentrations:
2SO3(g) ↔ 2SO2(g) + O2(g)
0.147-2X      2x               X
∴ [ O2] = 0.17 M / 4.5 L = 0.0378 M
So X = 0.0378
∴ [SO3] = 0.147 - 2 * 0.0378 =  0.0714 M
and [ SO2] = 2* 0.0378 = 0.0756 M
∴ by substitution in Kc formula 
∴Kc = [0.0756]^2 * [0.0378] / [0.0714] = 0.003

Sorry i dont understand the question can you rephase it?