1. Key idea: The $\text{C–OH}$ bond length decreases when the carbon attached to $\text{OH}$ has more double-bond character.

2. Methanol

   • In methanol, the bond is $\text{C(sp}^3\text{)–O}$.
   • There is no resonance involving this $\text{C–O}$ bond.
   • So it is a normal single bond, hence relatively longest.

3. Phenol

   • In phenol, the $\text{C–OH}$ bond is between oxygen and an $\text{sp}^2$ carbon of the benzene ring.
   • Lone pair on oxygen participates in resonance with the aromatic ring: $\text{Ar–OH} \leftrightarrow \text{Ar=O}^{+}\text{H}$
   • Thus the $\text{C–O}$ bond gets partial double-bond character, making it shorter than in methanol.

4. $p$-Ethoxyphenol

   • This has another substituent, $-\text{OC}_2\text{H}_5$, at the para position.
   • The ethoxy group is a $+M$ (electron-donating by resonance) group.
   • It donates electron density into the ring, which reduces the tendency of the phenolic oxygen lone pair to delocalize strongly into the ring.
   • Therefore, the phenolic $\text{C–OH}$ bond in $p$-ethoxyphenol has slightly less partial double-bond character than in phenol.
   • So its $\text{C–OH}$ bond is longer than phenol, but still shorter than methanol.

5. Increasing order of $\text{C–OH}$ bond length: $\text{phenol} < p\text{-ethoxyphenol} < \text{methanol}$

6. Check options:

   • A: $\text{phenol} < \text{methanol} < p\text{-ethoxyphenol}$ ❌
   • B: $\text{methanol} < p\text{-ethoxyphenol} < \text{phenol}$ ❌
   • C: $\text{methanol} < \text{phenol} < p\text{-ethoxyphenol}$ ❌
   • D: $\text{phenol} < p\text{-ethoxyphenol} < \text{methanol}$ ✅

Therefore, the correct option is D.