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Have you tried drawing them?
For this kind of questions, the way to go is to draw both conjugate bases, and see which is more stable.
The couple with the more stable base will have the lower pKa.
When we talk about acidity, in this case were referring to the ability for a species to dispense/lose a H+ ion from its structure. So, if we look at any species with an '-OH' hydroxyl group, the O-H bond is highly polarised in favour of the oxygen atom. Thus, the shared electrons that from the bond are susceptible of by 'taken' or localised onto the oxygen atom, with the hydrogen atom, having lost one of its electrons to oxygen, now free from the molecule. So, the oxygen will obtain a negative charge (gains hydrogen's shared electron) and the hydrogen will obtain a positive charge.
The structures of both phenol (C6H5OH) and ethanol (CH3CH2OH) are shown above. Whilst phenol is essentially a substituted benzene compound (i.e. an OH group has replaced one of the H atoms in benzene (C6H6)) and ethanol is a substituted alkane species, phenol contains delocalisation of electron density within the ring structure.
I'm not sure whether you are aware of this phenomenon or not, but essentially it's to do with the fact that the 6 carbon atoms each have one p-orbital electron which can be used to form 3 pi-bonds with a neighboring atom (as shown above). However, the electrons used to form these bonds are not localised between two carbon atoms, but rather shared/ distributed throughout all the carbons involved in the ring structure. This can be represented by a series of resonance structures which, when combined, provide a true description as to the electron density distribution in the molecule. In a structure such as an alcohol, this does not occur as there is no delocalised pi-bonding network.
For this question, start off with the phenol and methanol molecules with a localised negative charge on the oxygen atom (these are the conjugate bases of the original species). The stronger an acid, the more stable its conjugate base and this stability arises from being able to move/share the localised the negative charge around the atom via conjugation/the drawing of resonance structures. Attempt this for the conjugate base of phenol (where this is possible) and it should be clearer as to why it is more stable than a completely localised charge in the conjugate base of ethanol, which is more prone to undergoing reprotonation with the H+ once more.