The following data is available from the question:

MS: M = 164 g/mol

EA: Calculates empirical formula = C₅H₆O (IR confirms O is present).
Which with MW from MS gives molecular formula = C₁₀H₁₂O₂ . From here we get the IHD = 5

IR: There are significant absorptions at 1710cm^-1 due to a C=O and at 1600cm^-1 due to aC=C, C-O possible due to bands at 1275 and 1100cm^-1. There is no OH (about 3500cm^-1).

13C nmr: The proton decoupled spectrum shows a total of 8 peaks indicating 8 types of C. Since we know the formula, we know that since there are 8C from the MF then there is some symmetry. By analysis of the chemical shifts, we have a C=O (probably an acid derivative) at 167 ppm, 4 types in the region 125-140 ppm is typical of aromatic C, 60 ppm is typical of a C-O and those at 22 and 14 ppm most likely from hydrocarbon.

1H nmr: First of all we have 5 types of H showing up. After this, it's a good idea to tabluate the information to make sure you get it all correctly matched up:
 

The most significant structural information in the H nmr data is that we have a disubstituted phenyl group, an ethyl group : -CH₂-CH₃ most likely as an ethoxy group: -O-CH₂-CH₃ and an isolated methyl group : -CH₃

The IR gave us the C=O which the C nmr suggests is an acid derviative, such as an ester rather than a ketone (typically > 200ppm).

So with all this information we have the following pieces: C=O, -O-CH₂-CH₂- , -CH₃ and C₆H₄-.

Altogther...

ethyl 4-methybenzoate. The -CH₂- is deshielded by O and coupled to the -CH₃, while the other -CH₃ is isolated and slightly deshielded by the phenyl ring. The number of types of ArC (4) and the coupling in the Ar region of the H nmr (7-8ppm) suggests the para substitution pattern.
 

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit.
You should be asking yourself : "Does my answer give me what the H-nmr shows ?"