The thing to remember here is that the mass number, #A#, gives you the number of protons and the number of neutrons present inside a given atom's nucleus.
#color(blue)(|bar(ul(color(white)(a/a)A = Z + "no. of neutrons" color(white)(a/a)|)))#
Here
#Z# - the atomic number of the atom
In your case, the #"X"^(3-)# anion is said to have a mass number equal to #14#. This tells you that this particular ion has a total of #14# nucleons, which as you know is a term used to describe a proton or a neutron, inside its nucleus.
You can thus say, using the above equation, that your anion has
#14 = Z + "no. of electrons"#
Now, you know that this anion has #10# electrons. The charge of an ion can be calculated by subtracting the number of electrons that surround the nucleus from the number of protons that are located inside the nucleus
#"charge" = Z - "no. of e"^(-)#
In your case, a #3-# charge tells you that this anion contains #3# additional electrons than a neutral atom of #"X"# would contain. This means that you have
#-3 = Z - 10 implies Z = 7#
You now know that an atom of #"X"# has #7# protons inside its nucleus. As a result, the number of protons located inside the nucleus of #"X"# will be
#color(green)(|bar(ul(color(white)(a/a)color(black)("no. of neutrons" = 14 - 7 = 7)color(white)(a/a)|)))#
Moreover, you can say that you're dealing with the nitrogen-14 isotope, #""^14"N"#, the most abundant of the two stable isotopes of nitrogen.
The #"X"^(3-)# anion is the nitride anion, #"N"^(3-)#.
