Answer
$\dfrac{1}{3x}$.
Work Step by Step
Use $81=3^4$.
$\sqrt[3]{\dfrac{3xy^2}{81x^4y^2}}=\sqrt[3]{\dfrac{3xy^2}{3^4x^4y^2}}$
Use $\dfrac{a^n}{a^m}=\dfrac{1}{a^{m-n}}$.
$=\sqrt[3]{\dfrac{1}{3^{4-1}x^{4-1}y^{2-2}}}\\
=\sqrt[3]{\dfrac{1}{3^{3}x^{3}y^{0}}}$
Simplify using $a^0=1$:.
$=\sqrt[3]{\dfrac{1}{3^{3}x^{3}\cdot 1}}$
$=\sqrt[3]{\dfrac{1}{3^{3}x^{3}}}$
$=\sqrt[3]{\left ( \dfrac{1}{3x}\right )^3}$
Simplify.
$=\dfrac{1}{3x}$
Hence, the correct answer is $\dfrac{1}{3x}$.
