Solution.

The base of the prism is an isosceles right triangle, its area $S = \frac{a^2}{4}$, perimeter $P = a(\sqrt{2} + 1)$, and the lateral surface area $S_6 = P H = a H (\sqrt{2} + 1)$, where $H$ is the height of the prism.

Since $S_6 = 2 S$, then $a H (\sqrt{2} + 1) = \frac{a^2}{2} ; H = \frac{a}{2(\sqrt{2} + 1)} = \frac{a(\sqrt{2} - 1)}{2}$, the volume of the prism $V = S H = \frac{a^3 (\sqrt{2} - 1)}{8}$.

Answer: $\frac{a^3 (\sqrt{2} - 1)}{8}$.