You gave us one equation:
#bb"(1)"color(white)(m) "2SO"_3"(g)" rarr "2SO"_2"(g)" + "O"_2"(g)"; DeltaH = "198 kJ"#
However, you also need the equation for the formation of #"SO"_2#.
#bb"(2)"color(white)(m)"S(s)" + "O"_2"(g)" → "SO"_2"(g)"; ΔH = "-296.1 kJ"#
From these, you must devise the target equation:
#bb"(3)"color(white)(m)color(blue)("2S(s)" + "3O"_2"(g)" rarr "2SO"_3"(g)"; Δ_text(rxn)H^@ = ?)#
The target equation has #"2S(s)"# on the left, so you double equation (2).
When you double an equation, you double its #ΔH#.
#bb"(4)"color(white)(m)"2S(s)" + "2O"_2"(g)" → "2SO"_2"(g)"; ΔH = "-592.2 kJ"#
Equation (4) has #"2SO"_2"(g)"# on the right, and that is not in the target equation.
You need an equation with #"2SO"_2"(g)"# on the left.
Reverse Equation (1).
When you reverse an equation, you reverse the sign of its # ΔH#.
#bb"(5)"color(white)(m)"2SO"_2"(g)" + "O"_2"(g)" → "2SO"_3"(g)"; DeltaH = "-198 kJ"#
Now, you add equations (4) and (5), cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their #ΔH# values.
This gives us the target equation (6):
#color(white)(mmmmmmmmmmmmmmmmmm)ul(Δ_text(rxn)H^@"/kJ")#
#bb"(4)"color(white)(m)"2S(s)" + "2O"_2"(g)" → color(red)(cancel(color(black)("2SO"_2"(g)"))); color(white)(mmll)"-592.2"#
#bb"(5)"color(white)(m)ul(color(red)(cancel(color(black)("2SO"_2"(g)"))) + "O"_2"(g)" → "2SO"_3"(g)");color(white)(mm)ul("-198"color(white)(m))#
#bb"(6)"color(white)(m)color(blue)("2S(s)" + "3O"_2"(g)" rarr "2SO"_3"(g)";color(white)(mmmm) "-790")#
#Δ_text(rxn)H^@ = "-790 kJ"#
