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*