Because the power is relatively low at $2$, it is possible to directly expand the expression: $$\begin{align*} & \left(\mid x\mid +\frac{1}{\mid x \mid} -2\right)^2 \\ \\=\ & \left(\mid x\mid +\frac{1}{\mid x \mid} -2\right)\left(\mid x \mid +\frac{1}{\mid x \mid} -2\right) \\ \\=\ & \mid x\mid ^2 + 1 - 2\mid x\mid + 1 +\frac{1}{\mid x \mid ^2} -\frac{2}{\mid x\mid}-2\mid x\mid - \frac{2}{\mid x\mid } + 4 \\ \\=\ & \mid x\mid ^2 - 4\mid x\mid +\boxed{6} - \frac{4}{\mid x\mid} + \frac{1}{\mid x\mid ^2}\end{align*}$$

Alternatively, this problem can also be solved using multinomial expansion. Every term after expansion is in the form of $$\sum_{a+b+c=2}\binom{2}{a, b, c}\mid x\mid^a\times \frac{1}{\mid x\mid^b}\times (-2)^c$$

A term will be constant if and only if $a=b$. There are only two such cases: $a=b=0$ and $a=b=1$. It follows that the result is $$\binom{2}{0,0,2}(-2)^2 + \binom{2}{1,1,0}(-2)^0 = \frac{2!}{0!\cdot 0!\cdot 2!}\cdot 2^2 + \frac{2!}{1!\cdot 1!\cdot 0!}\cdot 2^0 =\boxed{6}$$