Let's start by showing the electron configuration of a neutral #"Fe"# atom, which has a total of **26** electrons.

#"Fe": 1s^(2)2s^(2)2p^(6) 3s^(2)3p^(6)4s^(2)3d^(6)#

#"Fe"^(3+)# has 3 less electrons than #"Fe"#; these electrons will be removed starting from the outermost shell, which in #"Fe"#'s case will be #4s# and #3d#. The first two will come from #4s# and the last one will come from #3d#. So,

#"Fe"^(3+): 1s^(2)2s^(2)2p^(6) 3s^(2)3p^(6)3d^(5)#

We've established that #"Fe"^(3+)# has #"26 - 23 = 26"# electrons. I assume the #"m"# you're asking for is actually #"m"_l#, the **magnetic** quantum number. Let's start counting the electrons that meet the #"n + l + m = 4"# criterion.

When #"n=1"#, you only get one choice for #"l"# and #""m"_l#, that is #"l = 0"# and #"m"_l = 0"#. This clearly doesn't satisfy the criterion. Moving on.

When #"n=2"#, you now get multiple values for #"l"# and #"m"_l#. If #"l"=0#, #"m"_l# is again equal to zero and your equation is #"2 + 0+ 0"#.

When #"l=1"#, #"m"_l# can be equal to #"-1"#, #0#, or #+1#. As you can see, we've found our first match: #"2 + 1 + 1 = 4#. Since the **spin quantum number** can be either #"-1/2"# or #"+1/2"#, we've found the first two electrons that match the equation.

When #"n=3"#, you've got multiple choices again, since #"l"# can be either #0#, #1# or #2#. When #"l = 0"#, #"m"_l =0# and we don't have a match.

Let's take the case when #"l=1"#. In this case, #"m"_l# can once again be #"-1"#, #0#, or #+1#. When #"m"_l# is equal to zero, we have another match: #"3 + 1 + 0 =4"#. Two more electrons meet the criterion.

If #"l"=2#, which means that #"m"_l# can be #"-2"#, #"-1"#, #"0"#, #"+1"#, and #"+2"#, we can see that we get one more option, #"m"_l "= -1"#. We have #"3 + 2 + (-1) = 4#.

Here's where it gets a little tricky. Notice that the #3d# subshell is only half-full. This means that *each of the five 3d-orbitals will only hold 1 electron*. As a result, just one electron meets the criterion this time.

Add up all the electrons and you'll get #"2 + 2 + 1 = 5"# electrons satisfy the criterion given.