A ketone gives a yellow precipitate when treated with I₂ in an alkaline solution. Thus, the ketone is 

Carbonyl groups that have a methyl group directly attached to the carbonyl carbon atom $\left(-\mathrm{CO}-{\mathrm{CH}}_3\right)$ give the iodoform test. In this test, I₂ with an alkali (NaOH) is treated with carbonyl compounds and the compounds that give the iodoform test give a yellow precipitate of CHI₃.

Among the given option, only methyl ketone gives the iodoform test as it has a methyl group that is directly attached to the carbonyl carbon atom.