To simplify $\left( \frac{3^{-2} \times 125^3}{6^3 \times 25^{-2}} \right)$:

1. Rewrite using prime factors:

   • $125 = 5^3$
   • $25 = 5^2$
   • $6 = 2 \times 3$

   Substituting these in, we have:

   $\frac{3^{-2} \times (5^3)^3}{(2 \times 3)^3 \times (5^2)^{-2}}$

   $= \frac{3^{-2} \times 5^9}{2^3 \times 3^3 \times 5^{-4}}$

2. Simplify:

   • For $3$, combine exponents: $3^{-2 - 3} = 3^{-5}$
   • For $5$, combine exponents: $5^{9 + 4} = 5^{13}$
   • For $2$, it's simply $2^3$

   $= \frac{5^{13}}{2^3 \times 3^5}$

The simplified form is $\frac{5^{13}}{2^3 \times 3^5}$.

So, the answer is (a): $\frac{3^5 \times 5^{13}}{2^3}$.