Getting the wrong pH value, according to data and online calculator
As you can see from the point that $n_\ce{OH} &gt; n_\ce{AcOH}$, there will be an excess of hydroxide ions. What you want to calculate is the pH for this amount of hydroxide ions within the total volume at this point.

Thus the calculation looks like:
\begin{align}
[\ce{OH-}] &amp;= \frac{n_\ce{NaOH}-n_\ce{AcOH}}{V_\text{a} + V_\text{b}}\\
\Leftrightarrow \mathrm{pOH} &amp;= -\lg ([\ce{OH-}])\\
\Leftrightarrow \mathrm{pH} &amp;= 14+\lg ([\ce{OH-}])
\end{align}

Your error is based on a rounding error ... please don't round intermediate result as all rounding errors will sum up. You just need to round the final results to a proper amount of decimal places.

The correct calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0024768 - 0.00246}{0.025 + 0.0258}\right) = 10.5194 \approx 10.52$$

vs. your calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0025 - 0.00246}{0.025 + 0.0258}\right) = 10.8962 \approx 10.90$$