Titration calculations

Just need to know if im stupid or this textbook is wrong:

25.0 cm3 of sulfuric acid were exactly neutralised by 28.5 cm3 of sodium hydroxide that had a
concentration of 0.750 mol dm-3. Calculate the concentration of the sulfuric acid.
H2SO4 + 2NaOH → Na2SO4 + H2O
In this question, we know the concentration of the NaOH, so we find the number of moles of this
first and then use the information gained to calculate the concentration of the sulfuric acid.
1 Number of moles of NaOH = concentration x volume
= 0.750 x 28.5/1000
= 0.0214 mol (to 3 sig.figs)
2 Ratio of H2SO4: NaOH from equation = 1:2
So, if we have 0.0214 moles NaOH, we’ll have 0.0107 moles of H2SO4.
3 Concentration of sulfuric acid = 0.0107/0.0285 = 0.375 mol dm-3

In step 3 shouldn't they divide by the volume of sulfuric acid NOT the volume of sodium hydroxide like whats going on here.