Let the radius of the circular base of the water tank be \(r\) meters. So, the area of the base = \(\pi r^2\) square meters. Given: \[ \pi r^2 = 616 \Rightarrow \frac{22}{7} \times r^2 = 616 \Rightarrow r^2 = \frac{616 \times 7}{22} = 196 \Rightarrow r = 14 \] Now, the height of the tank \(h = 21\) meters. Therefore, the total surface area of the tank (including the lid and the curved surface) is: \[ = 2\pi r^2 + 2\pi rh = 2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 21) = 2 \times \frac{22}{7} \times 14 \times 35 = 3080 \text{ square meters} \] So, the total surface area of the tank is 3080 square meters.