Reaction is:
${C_6}{H_5}{N_2}Cl\xrightarrow[{HCl}]{{CuCl}}{C_6}{H_5}Cl$
A. Gattermann’s reaction
B. Sandmeyer’s reaction
C. Wurtz’s reaction
D. Frankland’s reaction

Hint: The given reaction is the type of substitution reaction, where the diazonium salt is formed by reacting aryl amine with sodium nitrite in presence of aqueous hydrochloric acid which is further converted to aryl halide.

Complete step by step answer:
Gattermann’s reaction: The Gattermann’s reaction is defined as the reaction used for the formation of aromatic rings. In this reaction benzene diazonium chloride is converted to chlorobenzene on treating with copper in presence of hydrochloric acid.
The example of Gattermann’s reaction is shown below.
\[{C_6}{H_5}N_2^ + C{l^ - }\xrightarrow[{HCl}]{{Cu}}{C_6}{H_5}Cl + {N_2}\]
In this reaction, benzyl diazonium salt is reacted with copper in presence of hydrochloric acid to benzyl chloride and dinitrogen.
Sandmeyer’s reaction: The Sandmeyer’s reaction is defined as a reaction where aryl halide is formed from aryl diazonium salt where copper halide is used as a catalyst in presence of hydrochloric acid.
The example of sandmeyer’s reaction is shown below.

${C_6}{H_5}N_2^ + C{l^ - }\xrightarrow[{HCl}]{{CuCl}}{C_6}{H_5}Cl$
In this reaction, benzyl diazonium salt is reacted with copper chloride and hydrochloric acid to form benzyl chloride.
Wurtz’s reaction: Wurtz’s reaction is a type of coupling reaction where two mole of alkyl halide is reacted with sodium metal in presence of dry ether to form a hydrocarbon alkane with side product compound containing sodium and halogen.
The example of Wurtz’s reaction is shown below.
$2C{H_3} - Cl + 2Na\xrightarrow{{dry\;ether}}C{H_3} - C{H_3} + 2NaCl$
In this reaction two mole of alkyl chloride is reacted with two mole of sodium in presence of dry ether to form alkane and two mole of sodium chloride.
Frankland’s reaction: Frankland's reaction is similar to Wurtz’s reaction. In Frankland’s reaction Zn is replaced by sodium.
The example of Frankland’s reaction is shown below.
$2C{H_3} - Cl + 2Zn\xrightarrow{{dry\;ether}}C{H_3} - C{H_3} + 2ZnCl$
In this reaction two mole of alkyl chloride is reacted with two mole of zinc in presence of dry ether to form alkane and two mole of zinc chloride.

So, the correct answer is Option B.

Note: The Gattermann’s reaction is similar to Friedel Craft reaction. The main product formed in Gattermann’s reaction and Sandmeyer’s reaction is the same.