The number of $$sp^2$$ hybrid orbitals in a molecule of benzene is:

We begin by recalling what an $$sp^2$$ hybrid orbital is. When an atom undergoes $$sp^2$$ hybridisation, one $$s$$ orbital mixes with two $$p$$ orbitals to form three equivalent $$sp^2$$ hybrid orbitals. Thus, a single $$sp^2$$-hybridised atom always possesses exactly three $$sp^2$$ hybrid orbitals.

Benzene has the molecular formula $$\mathrm{C_6H_6}$$. Every carbon atom in benzene is $$sp^2$$ hybridised because each carbon forms two $$\sigma$$ bonds with neighbouring carbons, one $$\sigma$$ bond with a hydrogen, and retains one unhybridised $$p$$ orbital to participate in the delocalised $$\pi$$ system.

So, for one carbon atom we have

$$\text{Number of }sp^2\text{ hybrid orbitals per carbon}=3$$

Now we multiply by the total number of carbon atoms present in benzene:

$$\text{Total }sp^2\text{ hybrid orbitals}=3 \times 6 = 18$$

Hydrogen atoms do not contribute any $$sp^2$$ orbitals because a hydrogen atom contains only a $$1s$$ orbital, which does not hybridise with $$p$$ orbitals.

Therefore, the molecule of benzene as a whole contains $$18$$ $$sp^2$$ hybrid orbitals.

Hence, the correct answer is Option C.