Step 1: Recall what nitration needs.
Nitration by \( HNO_3 \) and \( H_2SO_4 \) generates the electrophile \( NO_2^+ \), and this electrophile attacks whichever ring has the highest electron density. So the fastest reacting compound is the one whose substituent is electron-donating, not electron-withdrawing.
Step 2: Sort the substituents by donating or withdrawing character.
The methyl group on toluene is an alkyl group, and alkyl groups donate electron density through hyperconjugation and the inductive effect, activating the ring. Fluorine and chlorine both pull electron density away from the ring inductively, and this withdrawal effect is stronger than their weak resonance donation, so both halogens deactivate the ring relative to benzene. The nitro group already on nitrobenzene withdraws electron density strongly by both induction and resonance, making that ring the most deactivated of all four.
Step 3: Rank the four compounds.
This places the compounds in the order toluene (most reactive), then fluorobenzene, then chlorobenzene, then nitrobenzene (least reactive), since activating groups speed up electrophilic attack while withdrawing groups slow it down.
Final Answer:
Toluene reacts fastest with the nitrating mixture because its methyl group is the only true electron donor among the four options.
\[ \boxed{\text{Toluene}} \]