The following data is available from the question:

MS: M = 198 g/mol (even, m and m+2, 1:1 ratio is the isotope pattern for 1 x Br)

EA: Elemental analysis data for C,H and N (i.e. standard analysis) showed 54.3% C and 5.57% H
With MW from MS gives the partial molecular formula = C₉H₁₁?  (MW = 119)
The E.A. was run for N but none was reported, but the MS shows Br therefore MF = C₉H₁₁Br  (MW = 119 + 79 =198)
From here we get the IHD = 4

IR: Not much, C-H above and below 3000cm^-1. There are significant absorptions at 1600cm^-1 due to C=C. There is no OH (about 3500cm^-1), no C=O around 1700cm^-1.

13C nmr: The proton decoupled spectrum shows a total of 7 peaks indicating 7 types of C. By analysis of the chemical shifts, we have 4 types in the region for Ar C 120-140 ppm, 50 and 47 ppm (slightly deshielded) and those at 26 ppm are most likely from a hydrocarbon portion.

1H nmr: First of all we have 4 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 most significant structural information.....monosubstituted aromatic = C₆H₅-,  a CH₃CH unit and a CH next to 5H
Summary....
The MS indicated MW =  198 g/mol and the presence of one Br atom.
The IR showed the presence of C=C (Ar C=C)
13C peaks at 120-140ppm suggests that there is a benzene ring, which is confirmed by the H NMR, and it's monosubstituted.
H nmr also gives a CH₃CH unit with the CH being next to 5H hence CH₃CHCH₂
Use this to check the molecular formula : C₉H₁₁Br = 9 x 12 + 11 x 1 + 1 x 79 = 198 g/mol
So with all this information we have the following pieces:

C₆H₅-,   CH₃CHCH₂ and Br

Altogether...

With the pieces we have :   C6H5-,  CH3CHCH2 and Br So we have a propane chain with two substituents, the phenyl and the Br. All we have to do is decide which is where.... Using the H nmr chemical shift of the CH group at 4.3ppm, we can conclude that it is the Br that is attached to the CH (more deshielded due to the electronegative Br).

2-bromo-1-phenylpropane

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 ?"

Note In the original examination, the CH2 signal was shown as a triplet (in reality it's even more complex than that if you look at it at higher frequencies). The unexpected issue here with the triplet is because the adjacent CH is a chirality center and the coupling of the CH₂ to the CH is not simple, because each H in the CH2 is different (diastereotopic) and each couple to the CH to give a doublet, so it's not a triplet we see, but really (more like) 2 doublets.