Step 1: Start with phenol and the first step.
Phenol heated with zinc dust loses its oxygen and is reduced to benzene. So X is benzene.
Step 2: Apply the chlorination step.
Benzene with $Cl_2$ in presence of $FeCl_3$ undergoes electrophilic aromatic substitution, putting a chlorine on the ring. So Y is chlorobenzene.
Step 3: Apply Friedel-Crafts acylation.
Chlorobenzene with $CH_3COCl$ and anhydrous $AlCl_3$ gains an acetyl group ($-COCH_3$) on the ring, forming a chloro acetophenone. This carbonyl is the key group for the next step.
Step 4: Apply the Wolff-Kishner reduction.
The reagent $N_2H_4/OH^-$ with glycol on heating reduces a carbonyl group ($C=O$) all the way to a $CH_2$ group. So the $-COCH_3$ becomes an $-CH_2CH_3$ (ethyl) group.
Step 5: Assemble the final molecule.
After reduction the ring carries an ethyl group (and the chlorine from earlier). The carbonyl is gone, leaving an ethyl substituted benzene type product.
Step 6: Match with the choices.
This final ethyl-bearing aromatic product matches Structure 3.
\[ \boxed{\text{Structure 3}} \]