Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "X" with 6-equivalents of "Y". "X" and "Y", respectively are

Borazine (B₃N₃H₆), also known as "inorganic benzene" due to its structural similarity to benzene, can be prepared from diborane and ammonia.

We start by writing the reaction for the standard preparation of borazine by heating diborane (B₂H₆) with ammonia (NH₃):

$$3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$$

The question states that borazine is prepared from 3 equivalents of "X" and 6 equivalents of "Y". From the balanced equation, 3 equivalents of $$B_2H_6$$ (diborane) correspond to X and 6 equivalents of $$NH_3$$ (ammonia) correspond to Y, which matches perfectly with the stoichiometry of the reaction.

Next, we consider why the other options do not apply. Option A, $$B(OH)_3 + NH_3$$, is not the standard method and does not match the stoichiometry.

Option C, $$B_2H_6 + HN_3$$, involves hydrazoic acid (HN₃), which is not used in borazine synthesis.

Option D, $$NH_3 + B_2O_3$$, while B₂O₃ can be used in some preparations, does not match the required stoichiometry of 3 equivalents.

Therefore, X = B₂H₆ and Y = NH₃, which corresponds to Option B.