Daily water consumption of 60,000 households

$V_c = 25(60,000) = 1,500,000 ~ \text{gal.}$

$V_c = 1,500,000 ~ \text{gal.} \times \dfrac{1 ~ \text{ft.}^3}{7.48 ~ \text{gal.}}$

$V_c = 200,535 ~ \text{ft.}^3$

Volume of 1 ft. depth of water in the reservoir

$V_r = A_b h$

$A_b = 1000 \times 43,560 = 43,560,000 ~ \text{ft.}^2$
$h = 1 ~ \text{ft.}$

$V_r = 43,560,000(1)$

$V_r = 43,560,000 ~ \text{ft.}^3$

Number of days

$N = \dfrac{V_r}{V_c}$

$N = \dfrac{43,560,000}{200,535}$

$N = 217 ~ \text{days}$ *answer*