The Wurtz-Fittig reaction involves condensation of:

First, let us recall the general idea behind coupling reactions that employ metallic sodium. The classic Wurtz reaction couples two molecules of an alkyl halide in the presence of sodium metal to give a symmetrical alkane:

$$2\,R{-}X \;+\; 2\,Na \;\rightarrow\; R{-}R \;+\; 2\,NaX$$

Here $$R$$ is an alkyl group and $$X$$ is a halogen such as $$Cl$$, $$Br$$ or $$I$$.

A modification of this reaction was developed to obtain compounds in which one carbon fragment is aromatic (aryl) and the other is aliphatic (alkyl). This modified reaction is called the Wurtz-Fittig reaction. In it, metallic sodium couples an aryl halide with an alkyl halide. Stating the general form first and then writing the equation, we have:

$$\text{Aryl halide} \;+\; \text{Alkyl halide} \;+\; 2\,Na \;\rightarrow\; \text{Alkyl-aryl hydrocarbon} \;+\; 2\,NaX$$

To make the discussion concrete, let the aryl halide be chlorobenzene $$C_6H_5Cl$$ and the alkyl halide be methyl chloride $$CH_3Cl$$. In dry ether and in the presence of two atoms of sodium, the reaction proceeds as:

$$C_6H_5Cl \;+\; CH_3Cl \;+\; 2\,Na \;\rightarrow\; C_6H_5CH_3 \;+\; 2\,NaCl$$

We see that one molecule of an aryl halide and one molecule of an alkyl halide jointly condense (couple) to yield an alkyl-substituted aromatic hydrocarbon. The presence of exactly these two different halides is what distinguishes the Wurtz-Fittig reaction from the ordinary Wurtz reaction (two alkyl halides) or other coupling variants.

Looking at the given options, the statement that matches this description is “one molecule of each of aryl-halide and alkyl-halide.”

Hence, the correct answer is Option B.