Find the value of the following expression:
$$\left[18-3\times \left\{3-3\times \left(3÷\frac{1}{3}-3\right)+3÷\frac{1}{3}+3\right\}-4\times 2\right]$$

To solve the expression, follow the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction):
First, simplify the innermost parentheses:

$$\left(3÷\frac{1}{3}-3\right)=(3\times 3-3)=(9-3)=6$$

Substitute this back into the curly braces and resolve the division and multiplication inside them:

$$\left\{3-3\times 6+3÷\frac{1}{3}+3\right\}$$

$$=\{3-18+(3\times 3)+3\}$$

$$=\{3-18+9+3\}$$

$$=\{15-18\}=-3$$

Now, substitute this result back into the outer square brackets:

$$\left[18-3\times (-3)-4\times 2\right]$$

$$=[18+9-8]$$

$$=[27-8]=19$$