Let $f\left(x\right)=x-2\sin x\cos x+\frac{1}{3}\sin 3x$

$\Rightarrow f^{\text{'}}\left(x\right)=1-2 \cos 2x+\cos 3x$

$\Rightarrow f^{"}\left(x\right)=4\sin 2x-3\sin 3x$

For maxima/minima $f^{\text{'}}\left(x\right)=0$

$\Rightarrow 1-2\left(2{\cos }^{2}x-1\right)+4{\cos }^{3}x-3\cos x=0$

$\Rightarrow (2 \cos x+\sqrt{3})(2 \cos x-\sqrt{3})(\cos x-1)=0$

$\cos x=\frac{-\sqrt{3}}{2},\frac{\sqrt{3}}{2},1$

$x=\frac{5\pi }{6},\frac{\pi }{6},0$

$f^{"}\left(\frac{5\pi }{6}\right)=-2\sqrt{3}-\sqrt{3}<0$

$f^{"}\left(\frac{\pi }{6}\right)=2\sqrt{3}-\sqrt{3}>0$

$f^{"}\left(0\right)=0$

So $x=\frac{5\pi }{6}$is local maxima point

Maximum value of $f\left(x\right)=f\left(\frac{5\pi }{6}\right)=\frac{5\pi }{6}+\frac{\sqrt{3}}{2}+\frac{1}{3}$

$=\frac{5\pi +2+3\sqrt{3}}{6}$

Hence this is the correct option.