$V = 15 ~ \text{yd.}^3 \times \left( \dfrac{3 ~ \text{ft.}}{1 ~ \text{yd.}} \right)^3$

$V = 405 ~ \text{ft.}^3$

$h = 10 ~ \text{ft.} ~ 9\frac{1}{2} ~ \text{in.} = 10 + \dfrac{9.5}{12}$

$h = 10\frac{19}{24} ~ \text{ft.}$

$V = A_bh$

$405 = A_b(1010\frac{19}{24})$

$A_b = 37.5289 ~ \text{ft.}^2$ *answer*