Solution: Note that \[ \omega = \frac{1-\sqrt{3}\iota}{2} \implies \omega \in \mathbb{Q} (\iota ,\sqrt{3} ). \] Therefore, \[ \mathbb{Q} (\iota ,\sqrt{3}, \omega ) = \mathbb{Q} (\iota , \sqrt{3} ). \] So, we need to determine the degree of the field extension $\mathbb{Q} (\iota ,\sqrt{3} )$. Since, $\iota \notin \mathbb{Q} (\sqrt{3} )$ and $\sqrt{3} \notin \mathbb{Q} (\iota )$, the polynomial \[ (x^2 + 1)(x^2 -\sqrt{3} ) \] is the minimal polynomial containing $\iota $ and $\sqrt{3} $. Therefore, the degree of the extension field will be the degree of the minimal polynomial, which is $4$.

This can also be viewed as \begin{align*} \deg \left( Q(\iota , \sqrt{3} , \omega ) \right) & = \deg \left( \mathbb{Q} (\iota ,\sqrt{3} ) \right) \\ & = \left[ \mathbb{Q} (\iota ,\sqrt{3} ): \mathbb{Q} \right] \\ & = \left[ \mathbb{Q} (\iota ,\sqrt{3} ): \mathbb{Q} (\iota ) \right] \left[ \mathbb{Q} (\iota ): \mathbb{Q} \right] \\ & = 2 \times 2 = 4. \end{align*}