To determine the number of species in which the central atom uses $$\mathrm{sp}^3$$ hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.

1. $$\mathrm{NH}_3$$ (Ammonia):

The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to $$\mathrm{sp}^3$$ hybridization.

2. $$\mathrm{SO}_2$$ (Sulfur dioxide):

The central atom is sulfur. Sulfur dioxide has 2 sigma bonds and 1 lone pair. Thus, the steric number is 3, which corresponds to $$\mathrm{sp}^2$$ hybridization.

3. $$\mathrm{SiO}_2$$ (Silicon dioxide):

The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to $$\mathrm{sp}$$ hybridization.

4. $$\mathrm{BeCl}_2$$ (Beryllium chloride):

The central atom is beryllium. Beryllium chloride has 2 sigma bonds and no lone pair. Thus, the steric number is 2, which corresponds to $$\mathrm{sp}$$ hybridization.

5. $$\mathrm{CO}_2$$ (Carbon dioxide):

The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to $$\mathrm{sp}$$ hybridization.

6. $$\mathrm{H}_2\mathrm{O}$$ (Water):

The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to $$\mathrm{sp}^3$$ hybridization.

7. $$\mathrm{CH}_4$$ (Methane):

The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to $$\mathrm{sp}^3$$ hybridization.

8. $$\mathrm{BF}_3$$ (Boron trifluoride):

The central atom is boron. Boron trifluoride has 3 sigma bonds and no lone pair. Thus, the steric number is 3, which corresponds to $$\mathrm{sp}^2$$ hybridization.

From the above analysis, the species in which the central atom uses $$\mathrm{sp}^3$$ hybrid orbitals are:

$$\mathrm{NH}_3$$
$$\mathrm{H}_2\mathrm{O}$$
$$\mathrm{CH}_4$$

Therefore, the number of species where the central atom uses $$\mathrm{sp}^3$$ hybrid orbitals is 3.