If the function $f(x)=x^3-6 x^2+a x+b$ satisfies Rolle's theorem in the interval $[1,3]$ and $f'\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then

$a=11$

Since f(x) satisfies conditions of Rolle's theorem.

∴  f(1) = f(3)

$\Rightarrow 1-6+a+b=27-54+3 a+b \Rightarrow a=11$

Now,

$f'(x)=3 x^2-12 x+a=3 x^2-12 x+11$

Clearly, $f'\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$. Hence, a = 11.