The following data is available from the question:

MS: M+ = 72 g/mol.  No Cl or Br  isotope pattern.

IR:  There is a strong absorptions at 1720 cm^-1 for a C=O.  There is no absorption at 3500cm^-1 for an -OH or -NH. No C=C band at  1600cm^-1.

13C nmr: The proton decoupled spectrum shows a total of 4 peaks indicating 4 types of C. By analysis of the chemical shifts, we have a very deshielded C at 209 ppm due to a C=O, most likely in a ketone or aldehyde (which are typically in the range 190-220 ppm). We have 3 other types at 39, 27 (slightly deshielded) and 8 ppm most likely from simple hydrocarbon components.

1H nmr: First of all we have 3 types of H showing up. After this, it's a good idea to tabulate the information to make sure you get it all correctly matched up:
 

The coupling patterns for the -CH₃ (1.1 ppm, a triplet) and the -CH₂- (2.4 ppm, a quartet) indicate that they are connected to each other to form an ethyl group : CH₃CH₂- .

Summary....

Simply, the H nmr data gives us a methyl group -CH₃ and an ethyl group CH₃CH₂-
The MS indicated MW = 60 g/mol.
The IR and 13C showed the presence of a ketone type C=O

So with all this information we have the following pieces: -CH₃,  -CH₂CH₃ , and C=O
Use this to check the molecular formula : C₄H₈O = 4 x 12 + 8 x 1 + 16 = 72 g/mol
This agrees with what was found in the MS.

Altogether...
 

With the pieces we have :  -CH3,  -CH2CH3 , and C=O  This means we have 2 ends ( -CH3 and -CH2CH3) and one middle piece, C=O which can only be out together in one way: Note the slightly deshielded chemical shift of the H adjacent to the C=O.

2-butanone

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 ?"