First find the concentration of the dilute sample.
Moles NaOH required to reach end point =cV
=0.0998 X 0.02243
=0.002239 moles of NaOH
As the NaOH reacts with the acetic acid in a 1:1 ratio,
Moles acetic acid also = 0.002239 moles
As the volume of acid used was 25mL
conc. of acid = n/V
=0.002239/0.025
=0.0895M acetic acid
As the original solution was diluted by a factor of ten, the concentration of the original is 0.895M