From the given list, the number of compounds with $$+4$$ oxidation state of Sulphur: $$SO_3, H_2SO_3, SOCl_2, SF_4, BaSO_4, H_2S_2O_7$$

We need to find compounds with sulphur in +4 oxidation state:

- $$SO_3$$: S is +6

- $$H_2SO_3$$: $$2(+1) + x + 3(-2) = 0$$, so $$x = +4$$. Yes.

- $$SOCl_2$$: $$x + (-2) + 2(-1) = 0$$, so $$x = +4$$. Yes.

- $$SF_4$$: $$x + 4(-1) = 0$$, so $$x = +4$$. Yes.

- $$BaSO_4$$: $$(+2) + x + 4(-2) = 0$$, so $$x = +6$$.

- $$H_2S_2O_7$$: This is pyrosulfuric acid. $$2(+1) + 2x + 7(-2) = 0$$, so $$2x = 12$$, $$x = +6$$.

Compounds with +4 oxidation state of S: $$H_2SO_3$$, $$SOCl_2$$, $$SF_4$$ = 3 compounds.

The answer is $$\boxed{3}$$.