The structure of the compound, which displays the following spectral data is IR = 1690, 1100 cm−1

1H NMR: δ 2.8(s, 3H), 3.8(s, 3H), 6.9(d, J = 8 Hz, 2H), 7.8(d, J = 8 Hz, 2H) 

13C NMR: δ 197, 165, 130, 129, 114, 56, 26

Concept:

Infrared spectroscopy: Infrared (IR) spectroscopy uses infrared radiation to study the vibrations of chemical bonds in a sample.

This technique is useful in identifying functional groups in molecules, as well as in analyzing the composition of organic and inorganic materials.

Explanation:

→  and have carbonyl ester group which have its peak at 1750cm^-1, which is not given in the option, thus option 1 and 2 are incorrect.
→ 
have four types of protons, thus satifies 1H NMR: δ 2.8(s, 3H), 3.8(s, 3H), 6.9(d, J = 8 Hz, 2H), 7.8

and, (d, J = 8 Hz, 2H) means hydrogens of two adjacent carbons are coupling at para position.

IR = 1690, 1100 cm−1 is for ketone group and c-o bond present in the structure.
→ 
have six types of protons thus do not satisfy peaks for proton NMR.

Conclusion:
So, The correct answer is option 3.