Here's what I got.
Your target solution must be
Now, your starting solution is
In order to reduce the concentration of the solution from
For any dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution will give you the dilution factor,
In this case, you have
#"DF" = (5 color(red)(cancel(color(black)(%))))/(2color(red)(cancel(color(black)(%)))) = color(blue)(2.5)#
The trick here is to realize that the dilution factor is also equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution.
#"DF" = V_"diluted"/V_"stock"#
This means that
#V_"diluted" = "DF" * V_"stock"#
In your case, you have
#V_"diluted" = color(blue)(2.5) * "10 mL" = "25 mL"#
So if the volume of the diluted solution is equal to
#"25 mL " - " 10 mL" = "15 mL"#
of water. In other words, if you add
I'll leave the answer rounded to two sig figs, but keep in mind that your values justify an answer rounded to one significant figure.