Answer
$B\lt C\lt A$
i.e. butanoic acid(compound marked as A) is the most soluble in water and 2-methyl-2-pentene is least soluble in water.
Work Step by Step
Butanoic acid ($CH_{3}CH_{2}CH_{2}COOH$) participates in H bonding with water molecule because here H is bonded with O. So, polarity of H is increased and as a result it will interact with $H_{2}O$ by H bonding. So, it is the most soluble among others.
2-methyl-2-pentene and 2-pentanone do not participate in H bonding with water molecule. But due to -CO- group present in 2-pentanone it will participate in dipole dipole interaction with polar $H_{2}O$ molecule.
On the other hand, no such interaction present in 2-methyl-2-pentene (Compound marked as B). So, Between compound B and C, C is more soluble in water than B. So, the least soluble in water is B.
