Let, the roots of the equation $3x^3-26x^2+52x-24=0$ are $\frac{a}{r}, a, ar.$

We know that, the sum and product of roots of the equation $A{x}^3+B{x}^2+Cx+D=0$ are respectively $-\frac{B}{A}$ and $-\frac{D}{A}.$

Hence, we have product of roots $\frac{a}{r}\times a\times ar=-\frac{\left(-24\right)}{3}$

$\Rightarrow a^3=8$

$\Rightarrow a=2.$

Now, the sum of the roots $\frac{a}{r}+a+ar=-\frac{\left(-26\right)}{3}$

$\Rightarrow \frac{2}{r}+2+2r=\frac{26}{3}$

$\Rightarrow \frac{2+2r+2r^2}{r}=\frac{26}{3}$

$\Rightarrow 6+6r+6r^2=26r$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow \left(3r-1\right)\left(r-3\right)=0$

$\Rightarrow r=\frac{1}{3}$ or $r=3.$

Hence, the roots are $\frac{2}{3}, 2, 6.$

And, the sum of two of them is $\frac{2}{3}+2=\frac{8}{3}$ or $2+6=8$ or $\frac{2}{3}+6=\frac{20}{3}.$

Out of these three possible sums $\frac{8}{3}$ is given in the option.