First determine the molarity of the sodium hydroxide solution.

4 g in 250 mL = 16 g/liter

molar mass of NaOH = 40 g/mole so you have 16/40 = 0.4 moles /L = 0.4 M solution.

Now if you take the produce of molarity * volume you get the number of moles of a substance in that volume.

If you write a balanced chemical equation for the reaction of NaOH + acetic acid you get:

NaOH + CH3COOH -->  CH3CooNa  + HOH

so for every mole of NaOH you neutralize one mole of acetic acid.

Since mL * M of the NaOH = mL * M for the acetic acid you can solve for the molarity of the acetic acid.

20 mL * 0.4 M = 25 mL * M of acetic acid

M = 20/25 * 0.4 = 0.32 M solution of acetic acid.