pH of a buffer made with boric acid and potassium dihydrogen borate The Problem


  Calculate the pH of a buffer that is $\pu{0.200M}$ $\ce{H3BO3}$ and $\pu{0.122M}$ $\ce{KH2BO3}$. The $K_\mathrm{a}$ for $\ce{H3BO3}$ is $7.3\times10^{-10}$


What I Tried

For each of these methods, I used $\pu{0.200M}$ as the concentration for the acid, $\ce{H3BO3}$, and $\pu{0.122M}$ as the concentration of its conjugate base, $\ce{H2BO3-}$.


Method one – Henderson–Hasselbalch equation 

$\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a}) = 9.13668$

${\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{0.122}{0.200}\right) = 9.13668 + (-0.21467) = 8.92201}$
Method two – Rice Table 
$$
\begin{array}{c|lcr}
 &amp; \text{$\ce{H3BO3}$} &amp; \text{$\ce{H3O+}$} &amp; \text{$\ce{H2BO3-}$} \\
\hline
\text{I} &amp; 0.200 &amp; 0 &amp; 0.122 \\
\text{C} &amp; -x &amp; +x &amp; +x \\
\text{E} &amp; 0.200-x &amp; x &amp; 0.122+x
\end{array}
$$
use x is small approximation
$K_\mathrm{a} = \ce{\frac{[\ce{H3O}][\ce{A-}]}{[\ce{HA}]}}$
$\ce{K_\mathrm{a} = \frac{0.122x}{0.200}}$
$x = 0.200 \times \frac{K_\mathrm{a}}{0.122}$
$x = 1.19672*10^{-9}$ – This equals the number of moles $\ce{H3O+}$
$\mathrm{pH} = -\log([\ce{H3O+}]) = 8.92201$  


Check $x$ is small approximation: $\frac{x}{0.122} = 9.081\times10^{-9}$ which is less than $5\ \%$, so the approximation is valid.

My question

The answer I got using both of these methods, $\mathrm{pH} = 8.92$, is not correct. The only conclusion I can come to is that I have missed some nuance of this question or I am making a careless mistake. Can somebody please tell me what I'm overlooking here?
Update: Per the advice of @DavePhD , I decided to trust my work and I approached my teacher with the problem. He just emailed me to say there must have been some error in the online system; 8.92 is the correct answer.

Thanks so much to everyone who took the time to help me with this!