The following data is available from the question:

MS: M+  at 114,  no isotope pattern for Cl or Br

EA: The % data when used together with the MW data available from the MS, gives the molecular formula = C₆H₁₀O₂ . From here we get the IHD = 2

IR: There are significant absorptions at 1730cm^-1 due to a C=O (which is higher than a typical ketone value of 1715cm^-1) and about 1600cm^-1 maybe due to a C=C. There are also strong bands at about 1200cm^-1 due to a C-O. It is also important to note the absence of bands for -OH, (so not an alcohol or carboxylic acid) especially given the O in the formula..

13C nmr: The proton decoupled spectrum shows 6 peaks indicating 6 types of C (therefore each is unique). By analysis of the chemical shifts, we get that 172 ppm is typical of O=C-O unit, 144 and100 ppm are 2 vinyl C (MF and IHD don't allow for them to be ArC), peaks at 40, 22 and 18 ppm are simple alkane type C and those at 36 ppm and 24 ppm are most likely just from slightly deshielded hydrocarbons.

1H nmr: There are 6 distinct 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:
 

Note the the integral total here can be matched to the number of H in the MF obtained from the MS and EA data.

From the coupling patterns one -CH₂- groups must be connected together as -CH₂CH₂- and the -CH₃ group is also conneected to a -CH₂- so we must have -CH₂CH₂CH₃.
We also get H₂C=CH- (note the complex coupling pattern typical of alkene units).

So with all this information we have the following pieces: -CH₂CH₂CH₃, O-C=O  and  H₂C=CH-, so we have all the pieces for the MF of C₆H₁₀O₂ .

From the pieces we have two ends and a middle, therefore it is only a question of which way around to put the middle piece....
H₂C=CH-O-C(=O)-CH₂CH₂CH₃, or H₂C=CH-C(=O)O-CH₂CH₂CH₃.
Which is it ?
The key is the chemical shift of the -CH₂- group, it is about 2.4 ppm which is more consistent with it being next to a C=O rather than directly attached to an O. Therefore the first structure is the answer.

Altogether...

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. In this question this would have allowed you to rule out all the other variations of the pieces.
You should be asking yourself : "Does my answer give me what the H-nmr shows ?"