Calculate the degree of unsaturation: the answer is 5. If the degree of unsaturation is 4 or greater, look for an aromatic ring, which has a degree of unsaturation of 4 (3 double bonds plus 1 ring). In addition to an aromatic ring, the molecule can have a carbon-carbon double bond, a carbonyl, or another ring.
Look for a carbonyl: the band at 1680 indicates a carbonyl. Another prominent band is the characteristic wide O-H stretch at about 3300. Note the small peaks to the left of 3000, indicating aromatic C-H stretch; aromaticity is supported by the abundance of peaks in the regions 1600-1585 and 1500-1400. C-O stretches show up from 1320-1000.
In the NMR, since we suspect a carboxylic acid, look for a broad singlet in the region 10-13.2 ppm.
Indeed, the broad peak at 10.2 indicates a carboxylic acid proton: -CO2H.
Note: an argument could be made that the molecule is an aldehyde. However, the presence of the wide O-H stretch band in the IR, the position of the carbonyl, and the absence of shoulder bands in the region 2830-2695 support that the molecule is a carboxylic acid rather than an aldehyde.
The aromatic region has 4 protons, indicating a disubstituted ring. Furthermore, the pattern in this region resembles para substitution (review the aromatic discussion). One group on the ring must be (or encompass) the -CO2H group; the other group on the ring is an alkyl group. The triplet at 4.0 ppm contains 2 protons and is shifted downfield relative to the peaks at 1.8 and 1.1 ppm. This triplet must be adjacent to an oxygen, since hydrogens on a carbon adjacent to an oxygen are shifted to 3.3-4.0 ppm. (If this triplet were adjacent to the aromatic ring, it would be at 2.2-3.0 ppm.)
Let's account for all 10 carbons: 6 in the ring, 1 in the carboxylic acid function. This leave 3 carbons for the alkyl group which is attached to the ring. We need to add one more carbon to the structure above:
The green hydrogens correspond to the sextet at 1.8 ppm (since they are next to carbons with a total of 5 hydrogens) and the purple hydrogens correspond to the triplet at 1.1 ppm (since they are next to a carbon with 2 hydrogens).
Exercise: Convince yourself that the following structures could NOT be the molecule represented in the given NMR. Hint: predict the NMR spectrum of each.