Oxidation state/number is the charge left on the central atom, when all the bonding pairs of electrons are broken with the charge assigned to the more electronegative atom. The sum of the oxidation numbers is always equal to the charge on the ion. Of course this procedure is a formalism; nevertheless, we can assign specific oxidation states/numbers for each of the given species.
#A.# We have sulfate dianion, #SO_4^(2-)#. Oxygen generally has an oxidation number of #-II#, and it does here. Thus #S_"ON"+4xx(-2)=-2; S_"ON"=VI+#
#B.# We have sulfurous acid, #H_2SO_3#. Again, oxygen generally has an oxidation number of #-II#, and it does here. Hydrogen generally has an oxidation number of #+I#, and it does here. Thus #S_"ON"+3xx(-2)+2xx(+1)=0; S_"ON"=IV+#. We would have got the same answer if we treated sulfurous acid as the adduct #H_2O*SO_2#; sulfur is clearly #S(+IV)# in #SO_2#. We could treat sulfuric acid also the water adduct of #SO_3#, i.e. #H_2O*SO_3#; clearly, again, sulfur is #S(+VI)# in #SO_3#
#C.# We have sulfide dianion, #S^(2-)#. The charge on the ion is #-2#, and the atom bearing the charge must have an oxidation number of #-II#.
#D.# Again we have sulfide dianion, #S^(2-)#, or at least #H_2S#. Hydrogen is #+I#, and the charge on the sulfur is #-2#, thus here #S# has an oxidation number of #-II#. Again, the sum of the oxidation numbers equal the charge on the ion; here #0#.
Can you tell me the oxidation numbers of #S# in #"thiosulfate"# anion, #S_2O_3^(2-)#? You could assign an average oxidation number, but I like to think that here each sulfur has distinct oxidation numbers, that is in thiosulfate, one sulfur atom has taken the place of oxygen in #SO_4^(2-)#, and so I would assign different oxidation numbers.
