A hydrocarbon $X\left(C_7{H}_{14}\right)$ on ozonolysis followed by work-up $Zn-H_{2}O$ gives $C_6{H}_{12}O$ as one of the product which on subsequent treatment with $KOH/I_2$ gives yellow precipitate and salt of a chiral acid. Hence, X could be

The correct option is B
Ozonolysis of $X\left(C_7{H}_{14}\right)$ gives $C_6{H}_{12}O$ which has one carbon less than X. Thus, X must contain double bond at terminal carbon.

Iodoform test: When Iodine and potassium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed.
Iodoform test is used to check the presence of carbonyl compounds with the structure $R-CO-C{H}_3$ or alcohols with the structure $R-CH\left(OH\right)-C{H}_3$ in a given unknown substance.
If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a $C{H}_{3}C=O$ group.
$C_6{H}_{12}O$ is a ketone with the structure $R-CO-C{H}_3$ which gives iodoform test as shown in the reaction.
Hence option (b) is correct.