The compound which shows metamerism is :

The correct option is C₄H₁₀O.

Metamerism is a type of structural isomerism that arises due to the presence of different alkyl chains (unequal distribution of carbon atoms) attached on either side of a polyvalent functional group (such as a divalent oxygen atom $-O-$, sulfur $-S-$, secondary amine $-\mathrm{NH}-$, or carbonyl group $-\mathrm{CO}-$).

Let us evaluate the options step-by-step:

1. C₃H₈O:
This molecular formula represents alcohols and ethers. The only ether possible with this formula is ethyl methyl ether:
${\mathrm{CH}}_3-O-{\mathrm{CH}}_2{\mathrm{CH}}_3$
Since we cannot write any other isomeric ether with a different distribution of alkyl groups around the oxygen atom, this compound cannot exhibit metamerism.

2. C₃H₆O:
This molecular formula represents aldehydes, ketones, or unsaturated ethers/alcohols. The only ketone possible is propanone (acetone):
${\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{CH}}_3$
Since there is only one ketone structure possible, it cannot show metamerism.

3. C₄H₁₀O:
This molecular formula represents saturated monohydric alcohols and ethers. Ethers possess a divalent oxygen atom ($-O-$) as their functional group. We can write multiple isomeric ethers with different alkyl groups on either side of the oxygen atom:
• Diethyl ether: ${\mathrm{CH}}_3{\mathrm{CH}}_2-O-{\mathrm{CH}}_2{\mathrm{CH}}_3$ (two ethyl groups)
• Methyl propyl ether: ${\mathrm{CH}}_3-O-{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3$ (one methyl and one propyl group)
• Methyl isopropyl ether: ${\mathrm{CH}}_3-O-\mathrm{CH}{\left({\mathrm{CH}}_3\right)}_2$ (one methyl and one isopropyl group)
These compounds differ only in the alkyl groups attached to the divalent oxygen atom. Thus, they are metamers, and C₄H₁₀O exhibits metamerism.