The following data is available from the question:

MS: M+  at 194,  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 = C₁₁H₁₄O₃ . From here we get the IHD = 5

IR: There are significant absorptions at 1730cm^-1 due to a C=O (which is higher than a typical ketone value of 1715cm^-1). 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 amount of O in the formula..

13C nmr: The proton decoupled spectrum shows 9 peaks indicating 9 types of C. Since we know the formula, there is some symmetry. By analysis of the chemical shifts, we get that 174 ppm is typical of O=C-O unit, 158-113 ppm are 4 x ArC, two peaks at 54 and 50 ppm are deshielded C and those at 36 ppm and 24 ppm are most likely just from slightly deshielded hydrocarbons. The off resonance decoupled 13C nmr shows two substituted ArC and 2 ArC-H (158-113 ppm), 2 -CH3 and 2 -CH2-.

1H nmr: There are 4 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.

The tough part here is deciding what the peaks at 3.7 ppm are all about. Is it a doublet or two singlets ? A couple of things can help us.... If its a doublet then we need a CH neighbour which itself would appear as a septet (or worse) for a CH coupled to two methyl groups. The there is no aliphatic CH in the off resonance decoupled 13C nmr and the two C atoms in these groups are DIFFERENT in the broad band 13C nmr.   THEREFORE it must be 2 singlets for two slightly different, deshielded methyl groups.

From the two doublets in the H nmr for the 4 ArH we can conclude that it is either ortho with two equivalent substituents or para with two different substituents. The 4 types of ArC suggest the later.

From the coupling patterns for the two -CH2- groups, they must be connected together as -CH2CH2- with no H on the next positions (which would complicate the coupling).

So with all this information we have the following pieces: -C6H4-, C=O, 2 x -OCH3  and  -CH2CH2-, so we have all the pieces for the MF of C₁₁H₁₄O₃ .

So we have the pieces, but its not over yet as there are many possible permutations of how to put them together.  Its the chemical shifts in the H and C nmr that provide the necessary key information to complete the question.  The -CH2CH2- chemical shifts are only 2.6 and 2.9 ppm, so they cannot be attached directly to O (since it would deshield and give shifts closer to 4 ppm).  This means it must be connected to the aromatic at one end and the C=O at the other as in -C₆H₄-CH₂CH₂-C=O only leaving us to add the two methoxy groups

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 ?"