The reaction of $$H_3N_3B_3Cl_3(A)$$ with $$LiBH_4$$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to $$H_3N_3B_3(Me)_3$$. Compounds (B) and (C) respectively, are:

We are told that compound $$A$$ has the molecular formula $$H_3N_3B_3Cl_3$$. In this six-membered $$B_3N_3$$ ring each boron atom is bonded to one chlorine atom, so the structure can be written as $$\left( BN\,BN\,BN \right)$$ with $$3$$ peripheral $$Cl$$ atoms.

First we let $$A$$ react with $$LiBH_4$$ in tetra­hydro­furan (THF). A well-known property of $$BH_4^-$$ is that it acts as a hydride donor; whenever it encounters a $$B\!-\!Cl$$ bond it delivers a hydride ion $$H^-$$ which replaces the chlorine. The purely ionic picture is

$$H_3N_3B_3Cl_3 + 3\,LiBH_4 \;\longrightarrow\; H_3N_3B_3H_3 + 3\,LiCl + 3\,BH_3.$$

All three chlorines are therefore substituted by hydrogens, giving $$H_3N_3B_3H_3$$. The compound $$H_3N_3B_3H_3$$ possesses an alternation of $$B$$ and $$N$$ atoms in a planar six-membered ring with delocalised $$\pi$$ bonding, exactly analogous to the $$\pi$$ system of benzene. Because of this analogy it is popularly called inorganic benzene, the systematic name being borazine.

Hence

$$B = \text{borazine} \; (\text{inorganic benzene}).$$

Next, compound $$A$$ is treated with another reagent $$C$$ to give $$H_3N_3B_3(Me)_3$$. We observe that every $$Cl$$ attached to boron has been replaced by a methyl group $$Me$$. To convert a $$B\!-\!Cl$$ bond into a $$B\!-\!Me$$ bond we need a reagent that can supply the carbanion fragment $$Me^-$$. The most common laboratory source of $$Me^-$$ is a Grignard reagent, specifically methylmagnesium bromide:

$$MeMgBr + B\!-\!Cl \;\longrightarrow\; B\!-\!Me + MgBrCl.$$

Applying this three times around the ring we have

$$H_3N_3B_3Cl_3 + 3\,MeMgBr \;\longrightarrow\; H_3N_3B_3(Me)_3 + 3\,MgBrCl.$$

Therefore

$$C = MeMgBr \;(\text{methylmagnesium bromide}).$$

Putting the two identifications together, compound $$B$$ is borazine and compound $$C$$ is methylmagnesium bromide. Examining the options, the pair (borazine, MeMgBr) corresponds to Option 4.

Hence, the correct answer is Option 4.