If $(2x+3y{)}^3-(2x-3y{)}^3=3y[A{x}^2+B{y}^2]$, then what is the value of $(A-B)$?

We can use the algebraic identity for the difference of cubes, $a^3-b^3=(a-b)(a^2+ab+b^2)$, where $a=2x+3y$ and $b=2x-3y$.
First, let's simplify $(a-b)$:
$$a-b=(2x+3y)-(2x-3y)=6y$$

Next, simplify $(a^2+ab+b^2)$:
$$a^2=(2x+3y{)}^2=4x^2+12xy+9y^2$$
$$b^2=(2x-3y{)}^2=4x^2-12xy+9y^2$$
$$ab=(2x+3y)(2x-3y)=4x^2-9y^2$$

Adding these terms together:
$$a^2+ab+b^2=(4x^2+12xy+9y^2)+(4x^2-9y^2)+(4x^2-12xy+9y^2)$$
$$=4x^2+4x^2+4x^2+12xy-12xy+9y^2-9y^2+9y^2$$
$$=12x^2+9y^2$$

Now substitute back into the formula:
$$(2x+3y{)}^3-(2x-3y{)}^3=6y(12x^2+9y^2)$$

We want to match it with the expression format $3y[A{x}^2+B{y}^2]$:
$$6y(12x^2+9y^2)=3y\cdot 2(12x^2+9y^2)=3y[24x^2+18y^2]$$

By comparing coefficients:
$$A=24$$
$$B=18$$

Thus, the value of $(A-B)$ is:
$$A-B=24-18=6$$