Step #1: Inspect the low mass region for immonium ions.
There are none visible.

Step #2: Inspect the low mass region for the b2 ion.
The b2 ion is 28 m/z heavier than the a2 ion. Peaks 185 and 213 are 28 m/z apart. Thus 213 is our b2 ion.

Step #3: Inspect the low mass region for the y1 ion.
This is a tryptic peptide, so there are two possibilities:
a) a peak at 147 indicates a y1 ion from a lysine.
or, b) a peak at 175 indicates a y1 ion from an argenine.

In this spectrum we have a peak at 175. This is our y1 ion. It indicates a C-terminal argenine.

Step #4: Inspect the high mass region for the y(n-1) ion.
The y(n-1) peak equals 1130 minus the b1 peak. However, we cannot see our b1 ion peak. But we see the b2 ion peak. We know the b2 ion is made of two amino acids, so we use table 4.2 to see what combinations of amino acids equals 213. Possibilities are:

D and P
or
I/L and V

We do not know the order, either, so we must try each amino acid.

Let's try D: 1130-115 = 1015; There is no peak at 1015, so this option cannot be true.
Now try P: 1130-97 = 1036; There is no peak at 1036, so this is not true either.
Now try I/L: 1130-113 = 1017; There IS a peak at 1016.6, which is ~1017. Thus 1016.6 is our y(n-1) ion peak. If we subtract the mass of valine from y(n-1), we get our y(n-2) ion that complements our b2 ion. 1016.6-99 = 917.6

We also know now that I/L is the first amino acid, and V is the second.

Step #5: Extend the y-ion series towards the lower m/z.
Step #6: Extend the b-ion series toward the higher m/z.
We left off at y(n-2). So look now for y(n-3). Find the difference between 917.5 and peaks to its left:
917.5 - 899.5 = 18 (too low to be an amino acid)
917.5 - 889.9 = 27.6 (still loo low)
917.5 - 856.3 = 61.2 (there is no amino acid with this mass)
917.5 - 842.3 = 75.2 (no amino acid has this mass)
917.5 - 832.2 = 85.3 (no amino acid has this mass)
917.5 - 788.4 = 129.1 This is the mass of E

So we think we found the y(n-3) ion peak. In most cases it will have a corresponding b3 ion peak. We find that by subtracting y(n-3) from the total mass. 1130 - 788 = 342. There is indeed a peak at 342. This is our b3 ion peak.

Notice that the difference between the b3 ion and the b2 ion is 342 - 213 = 129. This is the mass of E.

We can now continue in the same method as above to find the y(n-4) ion, y(n-5) ion, and so on.
IonPeak IonPeakAmino Acid
y(n-4)788.4 - 675.4 = 113 b41130 - 675 = 455I/L
y(n-5)675.4 - 546.3 = 129.1 b51130 - 546 = 584E
y(n-6)546.3 - 417.3 = 129 b61130 - 417 = 713E
y(n-7)417.3 - 288.3 = 129 b71130 - 288 = 842E
y(n-8)288.3 - 175.1 = 113.2 b81130 - 175 = 955I/L

Step #7: Calculate the mass of the proposed peptide sequence.
Our sequence is I/L-V-E-I/L-E-E-E-I/L-R
This is 113 + 99 + 129 + 113 + 129 + 129 + 129 + 113 + 175 = 1129.
We seem one dalton short. However, it says that [M+H] = 1130. So far we only calculated M. Thus we must add the mass of H. 1129 + 1 = 1130. We are good.

Step #8: Reconcile the amino acid content with the spectrum.
Here is where you would compare immonium ion peaks to the sequence you calculated. Since there are no immonium ion peaks visible, we move on to step 9.

Step #9: Attempt to identify all ions in the spectrum.
Small peaks might just be caused by noise, but large peaks are there for a reason. As you can see, there are a few additional large peaks. The can be explained here:
556 = doubly charged and loss of water (1130+1-18)/2 899.5 = 917.5 - 18 (loss of water)

Here is the spectrum with its peaks labeled:


The last step is to find the exact sequence. The easiest way to do this is to BLAST each possible sequence:
LVELEEELR
LVELEEEIR
LVEIEEELR
LVEIEEEIR
IVELEEELR
IVELEEEIR
IVEIEEELR
IVEIEEEIR

When you try the 5th sequence above, you get an exact match.