10. As shown in Figure 2, the volume of tetrahedron $D A B C$ is $\frac{1}{6}$, and it satisfies $\angle A C B=45^{\circ}$, $A D+B C+\frac{A C}{\sqrt{2}}=3$.
Then $C D=$ . $\qquad$
10. $\sqrt{3}$.

From $\frac{1}{3} A D \cdot\left(\frac{1}{2} B C \cdot A C \sin 45^{\circ}\right)$
$\geqslant V_{\text {tetrahedron DABC }}=\frac{1}{6}$,
that is, $A D \cdot B C \cdot \frac{A C}{\sqrt{2}} \geqslant 1$.
Also, $3=A D+B C+\frac{A C}{\sqrt{2}} \geqslant 3 \sqrt[3]{A D \cdot B C \cdot \frac{A C}{\sqrt{2}}} \geqslant 3$,

the equality holds if and only if $A D=B C=\frac{A C}{\sqrt{2}}=1$.
At this time, $A B=1, A D \perp$ plane $A B C$, hence $D C=\sqrt{3}$.