$\dfrac{a}{150 ~ \text{ft.}} = \dfrac{\frac{1}{4} ~ \text{in.}}{1 ~ \text{ft.}}$

$a = 37.5 ~ \text{in.} = 3.125 ~ \text{ft.}$

$r = 55 - a = 55 - 3.125$

$r = 51.875 ~ \text{ft.}$

Total volume (water + concrete)

$V_t = \text{Frustum of a cone}$

$V_t = \frac{1}{3}\pi h(R^2 + r^2 + Rr)$

$V_t = \frac{1}{3}\pi (150)[ \, 55^2 + 51.875^2 + 55(51.875) \, ]$

$V_t = 1,346,037.462 ~ \text{ft.}^3$

Volume of water

$V_w = \text{Cylinder}$

$V_w = \pi (50^2)(150)$

$V_w = 1,178,097.245 ~ \text{ft.}^3$

Volume of concrete

$V_c = V_t - V_w$

$V_c = 1,346,037.462 - 1,178,097.245$

$V_c = 167,940.217 ~ \text{ft.}^3$

Weight of water

$W_w = \gamma_w V_w = 62.4(1,178,097.245)$

$W_w = 73,513,268.09 ~ \text{lb}$

Weight of concrete

$W_c = \gamma_c V_c = 150(167,940.217)$

$W_c = 25,191,032.55 ~ \text{lb}$

Total weight when full of water

$W_t = W_w + W_c = 73,513,268.09 + 25,191,032.55$

$W_t = 98,704,300.64 ~ \text{lb} \times \dfrac{1 ~ \text{short ton}}{2000 ~ \text{lb}}$

$W_t = 49,352.15 ~ \text{short tons}$ *answer*