6. As shown in Figure 1, in the tetrahedron $P-ABC$, $PA \perp$ plane $ABC$, $\angle ABC=120^{\circ}$, $PA$ $=4$. If the radius of the circumscribed sphere of the tetrahedron $P-$ $ABC$ is $2 \sqrt{2}$, then the tangent value of the angle formed by line $PC$ and plane $ABC$ is $\qquad$
6. $\frac{2 \sqrt{3}}{3}$.

As shown in Figure 3, let $O_{1}$ be the circumcenter of $\triangle ABC$, and $O$ be the center of the circumsphere of the tetrahedron $P-ABC$.

Given that $PA \perp$ plane $ABC$, $OO_{1} \perp$ plane $ABC$, we know that $PA \parallel OO_{1}$.
Take the midpoint $D$ of $PA$.
Since $OP = OA = 2\sqrt{2}$, we know that quadrilateral $DAO_{1}O$ is a rectangle.

Also, $PA = 4$, so $OO_{1} = AD = 2$, and the radius $r$ of the circumcircle of $\triangle ABC$ is $O_{1}A = 2$.
In $\triangle ABC$, by $2r = \frac{AC}{\sin \angle ABC}$, we get
$AC = 2 \times 2 \sin 120^{\circ} = 2\sqrt{3}$.
Thus, $\tan \angle PCA = \frac{PA}{AC} = \frac{4}{2\sqrt{3}} = \frac{2\sqrt{3}}{3}$.

Therefore, the required value is $\frac{2\sqrt{3}}{3}$.