The compound which shows metamerism is :

Metamerism is a type of structural isomerism that occurs when compounds contain the same functional group (such as –O–, –CO–, –NH–, etc.) and have the same overall molecular formula, but the alkyl groups attached to the functional group are different on the two sides.

To decide which of the given formulas can show metamerism, we examine each one and ask whether more than one constitutional isomer with the same functional group can be drawn.

$C₃H₆O$

With this formula an aldehyde (propionaldehyde) or a ketone (acetone) can be formed, but the functional group (C=O) is not the same in the two possibilities (aldehyde vs ketone). Therefore no pair of metameric compounds exists for $C₃H₆O$.

$C₅H₁₂$

This is a saturated hydrocarbon (alkane). Alkanes contain only C–C and C–H bonds and have no functional group that can give rise to metamerism. Hence $C₅H₁₂$ cannot display metamerism.

$C₃H₈O$

Possible isomers are propanol (an alcohol) and methoxy‑ethane (an ether). The alcohol contains the –OH functional group, while the ether contains –O–. Because the functional groups differ, the two compounds are not metameric.

$C₄H₁₀O$

From this molecular formula we can draw two different ethers that have the same functional group (–O–) but different alkyl groups on either side:

• Methyl propyl ether: $CH₃–O–CH₂CH₂CH₃$ (methyl group on one side, propyl group on the other).
• Diethyl ether: $CH₃CH₂–O–CH₂CH₃$ (ethyl group on both sides, but the overall arrangement of alkyl groups differs from the first ether).

Both compounds have the identical molecular formula $C₄H₁₀O$ and share the same functional group (ether, –O–), yet the distribution of alkyl groups around the oxygen atom is different. This satisfies the definition of metamerism.

Since the other formulas cannot produce such a pair of isomers, the only compound that shows metamerism among the options is $C₄H₁₀O$.