The thing to remember here is that in order to have
What this tells you is that every time you have
In your case, you know that the sample contains
More specifically, the sample contains
#0.35 color(red)(cancel(color(black)("moles Cu"))) * overbrace((6.022 * 10^(23)color(white)(.)"atoms Cu")/(1 color(red)(cancel(color(black)("mole Cu")))))^(color(blue)("Avogadro's constant")) = color(darkgreen)(ul(color(black)(2.1 * 10^(23)color(white)(.)"atoms Cu")))#
The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of copper.