6. 9 .

By the binomial theorem, we have
$$
\begin{array}{l}
a-b \sqrt{2}=(1-\sqrt{2})^{2015} . \\
\text { Hence } a b=\frac{1}{4 \sqrt{2}}\left[(a+b \sqrt{2})^{2}-(a-b \sqrt{2})^{2}\right] \\
=\frac{1}{4 \sqrt{2}}\left[(3+2 \sqrt{2})^{2015}-(3-2 \sqrt{2})^{2015}\right] . \\
\text { Let } x_{n}=\frac{1}{4 \sqrt{2}}\left[(3+2 \sqrt{2})^{n}-(3-2 \sqrt{2})^{n}\right] .
\end{array}
$$

Then $x_{1}=1, x_{2}=6$,
$$
x_{n}=6 x_{n-1}-x_{n-2}(n \geqslant 3) \text {. }
$$

Thus, the units digit of $\left\{x_{n}\right\}$ are
$$
1,6,5,4,9,0,1,6,5, \cdots \text {. }
$$

Therefore, $x_{n+6} \equiv x_{n}(\bmod 10)$.
Hence $x_{2015} \equiv x_{5} \equiv 9(\bmod 10)$.
6. Let $a$ and $b$ be positive integers, and
$$
a+b \sqrt{2}=(1+\sqrt{2})^{2015} \text {. }
$$

Then the units digit of $a b$ is $\qquad$