Volume of the prism

The central angle of the regular octagon $\theta = \dfrac {360^\circ}8$ . Let the radius of the circumcircle of the octagon be $r$ . Then $r = \dfrac 4{\cos \frac \theta 2}$ . And:

• The area of a central triangle $A_\triangle = \dfrac 12 \left(\dfrac {4}{\cos \frac \theta 2}\right)^2 \sin \theta$
• The area of the octagon $A_\text{oct} = 8A_\triangle = \dfrac {4^3\sin \theta}{\cos^2 \frac \theta 2}$
• The volume of the prism $V = l A_\text{oct} = \dfrac {8 \times 4^3 \times 2 \sin \frac \theta 2\cos \frac \theta 2}{\cos^2 \frac \theta 2} = 4^5 \tan \frac \theta 2 = 1024 \tan 22.5^\circ \approx \boxed{424 \text{ m}^3}$

Consider the diagram.

$\tan 22.5=\dfrac{x}{4}$

$x=4\tan 22.5$

$2x=8\tan 22.5$

$Area~of~the~base=8\left(\dfrac{1}{2}\right)(4)(8\tan 22.5)\approx 53.019$

Therefore the volume is $53.019(8)\approx$ $\boxed{424}$