(B) 4

The field extension in question is \(Q(\sqrt{2}, \sqrt{3})\) over \(Q\). To find the degree of the extension, we consider the minimal polynomial of the elements being adjoined.

Both \(\sqrt{2}\) and \(\sqrt{3}\) are algebraic numbers, and their minimal polynomials over \(Q\) are \(x^2-2\) and \(x^2-3\), respectively. These polynomials are irreducible over \(Q\), so the degrees of the extensions \(Q(\sqrt{2})\) and \(Q(\sqrt{3})\) over \(Q\) are both 2.

Now, consider the extension \(Q(\sqrt{2})(\sqrt{3})\). Since \(\sqrt{3}\) is not an element of \(Q(\sqrt{2})\), the degree of the extension \(Q(\sqrt{2}, \sqrt{3})\) over \(Q(\sqrt{2})\) is 2 .

By the tower law, the degree of the extension \(Q(\sqrt{2}, \sqrt{3})\) over \(Q\) is the product of the degrees of the extensions \(Q(\sqrt{2})\) over \(Q\) and \(Q(\sqrt{2}, \sqrt{3})\) over \(Q(\sqrt{2})\). Hence, the degree is \(2 \times 2=4\)