Since the compound $\mathrm{X}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}ight)$ gives iodoform test, it must have $-\mathrm{COCH}_{3}$ groups hence it should be $\mathrm{CH}_{3} \mathrm{COCH}_{3} .$ Therefore $\mathrm{X}$ should $\mathrm{be} \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} .$
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