The chemical equation is
"2SO"_3 ⇌ "2SO"_2 + "O"_2
The initial amount of "SO"_3 is
0.80 color(red)(cancel(color(black)("g SO"_3))) × ("1 mol SO"_3)/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "0.009 99 mol SO"_3
The volume of the container is 1 L, so the initial concentration of "SO"_3 is
["SO"_3] = "0.009 99 mol"/"1 L" = "0.09 99 mol/L"
We can setup an ICE table to solve this problem.
color(white)(mmmmmmmm)"2SO"_3 color(white)(l)⇌ color(white)(l)"2SO"_2 + "O"_2
"I/mol·L"^"-1": color(white)(mm)"0.009 99"color(white)(mmm)0color(white)(mmm)0
"C/mol·L"^"-1":color(white)(mmll) "-2"xcolor(white)(mmmll)"+2"xcolor(white)(mll)"+"x
"E/mol·L"^"-1": color(white)(m)"0.009 99 - 2"xcolor(white)(ml)2xcolor(white)(mmll)x
We know that 80 % of the "SO"_3 has decomposed.
∴ The concentration remaining at equilibrium is
"0.20 × 0.009 99 = 0.002 00"
∴ "0.009 99 -"color(white)(l)2x = "0.002 00"
2x = "0.009 99 - 0.002 00" = "0.007 99"
x = "0.007 99"/2 = "0.004 00"
So,
["SO"_3]_text(eq) = "0.002 00 mol/L"
["SO"_2]_text(eq) = 2x color(white)(l)"mol/L" = "0.007 99 mol/L"
["O"_2]_text(eq) = x color(white)(l)"mol/L" = "0.004 00 mol/L"
K_c = (["SO"_2]^2["O"_2])/(["SO"_3]^2) = (("0.007 99")^2 × "0.004 00")/("0.002 00")^2 = 0.064
