Firstly, there is an error in the calculations done so far.
In the step where the Average volume of oxalic acid was calculated, it gets to the average of 12.2 mL ± 0.410%. This is ok. But the problem is converting from mL to dm³.
1 mL is equal to 10⁻³ L and L is the same as dm³.
So
1 mL --- 0.001 dm³
12.2 mL --- x
[tex]\begin{gathered} \frac{1}{12.2}=\frac{0.001}{x} \\ x=12.2\cdot0.001 \\ x=0.0122 \end{gathered}[/tex]
That is, the result should be 0.0122 dm³ ± 0.410%.
Following the next steps, we would have
[tex]n=(0.10018\pm0.07918\%)\cdot(0.0122\pm0.410\%)=0.0012222\pm0.48918\%[/tex]
And:
[tex](0.0012222\pm0.48918\%)\cdot2=0.0024444\pm0.48918\%[/tex]
Finally, to get the concentration of NaOH, we divide this by the volume of the solution of NaOH, which is 25 mL ± 0.03 mL = 25 mL ± 0.12% = 0.025 dm³ ± 0.12%:
[tex]C=\frac{n}{V}=\frac{0.0024444mol\pm0.48918\%}{0.025dm^3\pm0.12\%}=0.097776mol\cdot dm^{-3}\pm0.60918\%[/tex]
