I assume you used M1V1=M2V2?

V₂=V₁ x C₁/C₂ is the formula I used.

Multiply both sides by C₂

http://www.chembuddy.com/?left=concentration-questions&right=molarity-dilution-q1

and

http://www.chembuddy.com/?left=concentration-questions&right=molarity-dilution-q2

Your answer is (almost) correct - to be precise you need 547.2mL (taking density changes into account).

Try CASC for such calculations.

Oops, sorry - somehow I managed to calculate correct result for percentage, but to not spot that equation you used is valid only for molar concentrations

To calculate properly amount of water you need to add you have to know density of the 30% and 7% solutions. Why? Because mass percentage is defined as weight of solute/weight of solution (see equation 1.1 in my concentration lectures).

Densities are usually given in tables (for NaOH you will find density table on my website too, in the CASC description). In your case - 30% solution has a density of 1.33 g/mL and 7% - 1.08 g/mL.

Your original solution has a mass of 125*1.33 = 166.3 g. In this solution there is 30/100*166.3 = 49.88 g NaOH. It will be 7% (or 0.07) of the mass of the final solution - so the final solution mass is 49.88/0.07 = 712.5 g.

Amount of water that have to be added is the difference between the final mass and the starting mass - 712.5-166.3 = 546.2 g. To be very precise you should take water density into account, but if you assume that 1 mL of water is just 1 g you will not make large error - so you have to add 546.2 mL of water (my previous answer was a little bit more precise, as it took into account more significant digits and the real water density).

Your original approach will not give correct result in case of mass percentages and concentrated solutions!