Here's what I got.
The trick here is to realize that when you're diluting a solution, the increase in volume must match the decrease in concentration
In other words, the volume will increase by a factor and the concentration will decrease by the same factor.
#color(blue)(ul(color(black)("DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"))) ->#the dilution factor
In your case, the concentration of the solution must decrease by a factor of
#"DF" = (3.02 color(red)(cancel(color(black)("M"))))/(0.150 color(red)(cancel(color(black)("M")))) = color(blue)(20.133)#
This means that the volume must have increased by the same dilution factor.
You will thus have
#color(blue)(20.133) = V_"diluted"/V_"concentrated" implies V_"concentrated" = V_"diluted"/color(blue)(20.133)#
Therefore, the student will need
#V_"concentrated" = "125 mL"/color(blue)(20.133) = color(darkgreen)(ul(color(black)("6.21 mL")))#
The answer is rounded to three sig figs.
The difference between the volume of the stock solution and the volume of the diluted solution represents the volume of water needed for the dilution.
Also, keep in mind that when you're dealing with strong acids and strong bases, you must add the acid or the base to water, not the other way around!