The acidic strength of the given compounds follows the order: 

$$\begin{gathered} \mathrm{I}) {\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{COOH} \\ \mathrm{II})\hspace{0.17em}{\mathrm{CH}}_3-\mathrm{O}-\mathrm{CH}=\mathrm{CH}-\mathrm{COOH} \\ \mathrm{III}) {\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH} \end{gathered}$$

Compound (III) is least acidic because of the presence of electron releasing groups. In conjugate base of compound (I), there is cross resonance because the two double bonds (-C=C and -C=O) and negative charge are in conjugation.

${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-{\mathrm{O}}^{-}$

In conjugate base of compound (II), cross resonance further enhanced because the two double bonds (-C=C and -C=O), lone pairs of O and negative charge are in conjugation.

${\mathrm{CH}}_3-\underset{..}{\overset{..}{\mathrm{O}}}-\mathrm{CH}=\mathrm{CH}-{\mathrm{COO}}^{-}$

Hence, order becomes I>II>III