6. In tetrahedron $ABCD$, $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. If a sphere passing through point $D$ is tangent to plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron with a radius of 1, then the radius of the circumscribed sphere of tetrahedron $ABCD$ is . $\qquad$
6. 3 .

As shown in Figure 2, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ at point $E$ and $DF \perp AC$ at point $F$.
$$
\begin{array}{l}
\text { Then } H E \perp A B, H F \perp A C, \text { and } \\
A E=A F=A D \cos 45^{\circ}=\sqrt{6} . \\
\text { By } \triangle A E H \cong \triangle A F H \\
\Rightarrow \angle H A E=30^{\circ} . \\
\text { Also } A H=\frac{A E}{\cos 30^{\circ}}=2 \sqrt{2}, \\
D H=\sqrt{A D^{2}-A H^{2}}=2,
\end{array}
$$

Thus, $D H$ is the diameter of a sphere with radius 1.
Therefore, the center $O$ of the circumscribed sphere of tetrahedron $ABCD$ lies on the extension of $D H$.
Let the radius of the circumscribed sphere be $r$. Then
$$
r^{2}=(r-2)^{2}+(2 \sqrt{2})^{2} \Rightarrow r=3 .
$$