Given that, $f(x) = \frac{5x+2}{3x-5}$
And, $g(x) = x^{2}-2x-1$
Now, $f(3) = \frac{5(3)+2}{3(3)-5} = \frac{17}{4}$
$\Rightarrow f(f(3)) = f\left(\frac{17}{4}\right)$
$\Rightarrow f\left(\frac{17}{4}\right) = \frac{5\left(\frac{17}{4}\right)+2}{{3\left(\frac{17}{4}\right)-5}}$
$\Rightarrow f\left(\frac{17}{4}\right) = \dfrac{\frac{85+8}{4}}{\frac{51-20}{4}}$
$\Rightarrow f\left(\frac{17}{4}\right) =\frac{93}{31} = 3$
$\Rightarrow g(f(f(3))) = g\left(f\left(\frac{17}{4}\right)\right) = g(3)$
$\Rightarrow g(3) = 3^{2} – 2(3)-1 = 9-6-1 = 9-7 = 2$
$\therefore$ The value of $g(f(f(3))) = 2.$
Correct Answer $:\text{A}$