$f\left(x\right)=x^3+b{x}^2+ax;f^{\mathrm{\prime }}\left(x\right)=3x^2+2bx+a$

Since Rolle’s theorem is verified

$\Rightarrow f^{\mathrm{\prime }}\left(c\right)=0\Rightarrow 3c^2+2bc+a=0$

$\Rightarrow c=\frac{-2b\pm \sqrt{4b^2-12a}}{6}\Rightarrow c=\frac{-b\pm \sqrt{b^2-3a}}{3}$

Given $C=2+\frac{1}{\sqrt{3}}; \therefore 2+\frac{1}{\sqrt{3}}=\frac{-b+\sqrt{b^2-3a}}{3}$

Comparing on both sides

$\Rightarrow \frac{-b}{3}=2\Rightarrow b=-6\Rightarrow \frac{\sqrt{b^2-3a}}{3}=\frac{1}{\sqrt{3}}$

$\Rightarrow b^2-3a=3$

$\Rightarrow \left(36\right)-3a=3\Rightarrow 3a=33\Rightarrow a=11$

$\therefore$the values of a and b are $11, –6$