The compound which shows metamerism is :

The correct answer is C₄H₁₀O.

Understanding Metamerism:
Metamerism is a type of structural isomerism shown by compounds sharing the same molecular formula but having different alkyl groups attached to the same polyvalent functional group (such as a divalent ether oxygen $-O-$, a divalent thioether sulfur $-S-$, a carbonyl group $-\mathrm{CO}-$, or an imino group $-\mathrm{NH}-$).

Let us evaluate each of the given options:

1. C₅H₁₂: This is an alkane (pentane). It has no heteroatom or polyvalent functional group, so it cannot exhibit metamerism.
2. C₃H₈O: This represents alcohols and ethers. The only ether possible with three carbon atoms is methoxyethane:
${\mathrm{CH}}_3-O-C_2{H}_5$
Since there is only one way to arrange the carbon atoms around the oxygen atom, it cannot form metamers.
3. C₃H₆O: This represents aldehydes, ketones, or unsaturated alcohols/ethers. The only ketone possible with three carbon atoms is propanone:
${\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{CH}}_3$
We cannot vary the alkyl groups on either side of the carbonyl group, so it does not show metamerism.
4. C₄H₁₀O: This molecular formula represents ethers and alcohols with four carbon atoms. We can write multiple ether structures by varying the alkyl groups attached to the divalent oxygen atom:
• Diethyl ether (ethoxyethane): $C_2{H}_5-O-C_2{H}_5$
• Methyl propyl ether (1-methoxypropane): ${\mathrm{CH}}_3-O-{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3$
• Methyl isopropyl ether (2-methoxypropane): ${\mathrm{CH}}_3-O-\mathrm{CH}{\left({\mathrm{CH}}_3\right)}_2$

Because these compounds have different alkyl chains distributed around the same functional group (-O-), they are metamers of each other. Therefore, the compound C₄H₁₀O shows metamerism.