The number of $$S - S$$ bonds in $$SO_3$$, $$S_2O_3^{2-}$$, $$S_2O_6^{2-}$$ and $$S_2O_8^{2-}$$ respectively are

The presence of an $$S-S$$ bond can be decided only after drawing the accepted Lewis / VSEPR structure of each anion or molecule. We therefore write the skeletal structures first and then count the links between the two sulphur atoms.

Case 1: $$SO_3$$ (sulphur trioxide)
The preferred structure is a trigonal-planar $$\ce{O=S(=O)2}$$ arrangement in which the single sulphur atom is $$\text{sp}^2$$ hybridised and is bonded only to oxygen. There is no second sulphur atom, hence
Number of $$S-S$$ bonds in $$SO_3 = 0$$.

Case 2: $$S_2O_3^{2-}$$ (thiosulphate ion)
The accepted structure is
$$\ce{{}^{-}O\!\!-\!S(=O)_2\!-\!S^{2-}}$$
One sulphur is in the centre (formal oxidation state +6) and a second sulphur (oxidation state −2) is attached directly to it, giving exactly one $$S-S$$ linkage. Therefore
Number of $$S-S$$ bonds in $$S_2O_3^{2-} = 1$$.

Case 3: $$S_2O_6^{2-}$$ (dithionate ion)
The structure is two $$\ce{SO_3^-}$$ groups joined through an $$S-S$$ bond:
$$\ce{^{-}O_3S\!\!-\!S O_3^{-}}$$
Because the two sulphur atoms are directly connected, there is exactly one $$S-S$$ bond. Hence
Number of $$S-S$$ bonds in $$S_2O_6^{2-} = 1$$.

Case 4: $$S_2O_8^{2-}$$ (peroxydisulphate ion)
This ion contains a peroxide $$O-O$$ bridge, not an $$S-S$$ link:
$$\ce{^{-}O_3S\!\!-\!O\!-\!O\!\!-\!S O_3^{-}}$$
The two sulphur atoms are separated by the $$O-O$$ bridge and never bonded directly. Therefore
Number of $$S-S$$ bonds in $$S_2O_8^{2-} = 0$$.

Collecting the results:
$$SO_3 : 0,\; S_2O_3^{2-} : 1,\; S_2O_6^{2-} : 1,\; S_2O_8^{2-} : 0$$

Thus the correct choice is
Option C which is: $$0, 1, 1, 0$$