Given $f_a\left(x\right)={\tan }^{-1}2x-3ax+7$ is non-decreasing function in $\left(-\frac{\pi }{6},\frac{\pi }{6}\right)$, so 

${f_a}^{\text{'}}\left(x\right)=\frac{2}{1+4x^2}-3a\ge 0$

i.e. $a\le {\left(\frac{2}{3\left(1+4x^2\right)}\right)}_{\min }$ 

Now, maximum value of $a$ is at $x=0$

i.e. $a_{\max }=\overline{a}=\frac{6}{9+{\pi }^2}$

Hence, $f_{\overline{a}}\left(\frac{\pi }{8}\right)={\tan }^{-1}\frac{\pi }{4}-3\left(\frac{6}{9+{\pi }^2}\right)\left(\frac{\pi }{8}\right)+7$

$={\tan }^{-1}\frac{\pi }{4}-\frac{9\pi }{4\left({\pi }^2+9\right)}+7=8-\frac{9\pi }{4\left({\pi }^2+9\right)}$