The thing to remember about dilutions is that you are decreasing the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.
This implies that if you increase the volume of a solution by a factor, let's say
#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))#
In your case, the concentration of the solution decreased by a factor of
#"DF" = c_"concentrated"/c_"diluted" = (15.0 color(red)(cancel(color(black)("M"))))/(3.00color(red)(cancel(color(black)("M")))) = color(blue)(5)#
which can only mean that its volume increased by a factor of
#V_"diluted" = color(blue)(5) xx "25.0 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("125 mL")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
So remember, the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution tells you the ratio that must exist between the volume of the diluted solution and the volume of the concentrated solution.
You can use proportions to do this calculation. We need a 1/5th reduction in concentration which implies that our final diluted volume should be 5x your initial volume of nitric acid.
The common formula used in chemistry basically illustrates conservation of the quantity of acid after dilution as there is no consumption of acid after dilution.
Where M is the molarity of the solution and V the volume.
Out of interested you might want to compare the concentration of this acid with commercially available nitric acid which is around 68%. That is around 10-11 M concentration.