Whether an alkyl halide undergoes substitution or elimination depends on the reagent and medium. In aqueous potassium hydroxide the hydroxide ion is heavily hydrated and behaves mainly as a nucleophile, attacking the carbon bearing bromine to give an alcohol by substitution. In alcoholic potassium hydroxide some ethoxide ion is generated, and the less-solvated, more basic species abstracts a beta-hydrogen, promoting elimination to give an alkene. The bromide-leaving option is wrong because the leaving group departs in both solvents; only the product differs. The temperature option is incorrect because the change is due to reagent behaviour, not to a large difference in boiling-driven temperature. The potassium oxidation-state option is wrong because potassium remains in its +1 state throughout and plays no redox role. The deciding factor is the relative solvation and basicity of the attacking species: water heavily solvates hydroxide and blunts its basicity so it prefers carbon attack, whereas in ethanol the partially formed ethoxide is a stronger, less-hindered base that prefers to grab a beta-hydrogen. This same competition between substitution and elimination is central to predicting products throughout haloalkane chemistry. This matches the NCERT contrast between aqueous and alcoholic KOH. Sanity check: nucleophilic substitution gives a C-OH bond while base-driven elimination gives a C=C bond, exactly the two observed products.