
Volume of pyramidal top
$V_p = \frac{1}{3}A_bh = \frac{1}{3}(\frac{4}{12})^2(\frac{2}{12})$
$V_p = \frac{1}{162} ~ \text{ft.}^3$
Volume of frustum part
$V_f = \frac{1}{3}(A_1 + A_2 + \sqrt{A_1A_2})h$
$V_f = \frac{1}{3}[ \, (\frac{6}{12})^2 + (\frac{4}{12})^2 + \sqrt{(\frac{6}{12})^2(\frac{4}{12})^2} \, ] (6 - \frac{2}{12})$
$V_f = \frac{665}{648} ~ \text{ft.}^3$
Volume of 1 post
$V_1 = V_p + V_f = \frac{1}{162} + \frac{665}{648}$
$V_1 = \frac{223}{216} ~ \text{ft.}^3$
Volume of 150 posts
$V = 150V_1 = 150(\frac{223}{216})$
$V = 154.86 ~ \text{ft.}^3$ answer