6. As shown in Figure 1, in the tetrahedron
$$
\begin{array}{l}
A B C D \text {, } D A=D B=D C \\
=2, D A \perp D B, D A \perp
\end{array}
$$
$D C$, and the cosine of the angle between $D A$ and the plane $A B C$ is $\frac{\sqrt{6}}{3}$. Then the radius $R$ of the circumscribed sphere of the tetrahedron is $\qquad$
6. $\sqrt{3}$.

As shown in Figure 3, draw DO $\perp$ plane $ABC$ at point $O$, connect $AO$ and extend it to intersect $BC$ at point $E$, and connect $DE$.

Thus, $\angle DAE$ is the angle between $DA$ and plane $ABC$,
$$
\cos \angle DAE = \frac{\sqrt{6}}{3}.
$$

From the given conditions, we know that $DA \perp$ plane $DBC$, $O$ is the circumcenter of $\triangle ABC$, and $AB = AC = 2\sqrt{2}$.
Therefore, $DA \perp DE$, and $E$ is the midpoint of side $BC$.
Combining with equation (1), we know that $AE = \sqrt{6}$.
Thus, $BE = \sqrt{AB^2 - AE^2} = \sqrt{2}$
$\Rightarrow BC = 2BE = 2\sqrt{2}, DB \perp DC$.
Hence, $DA, DB, DC$ are mutually perpendicular, and the radius $R$ of the circumscribed sphere of the tetrahedron is $\sqrt{3}$.