Hello Cat,
To find the total consumption $T$ of water for 24 hours, we may state:
$$T=\int_{0}^{24} C(t)\,dt=25\int_{0}^{24} e^{-\frac{(t-15)^2}{20}}\,dt$$
Now, the integrand in this problem does not have an anti-derivative expressible in elementary terms, so we must use either the error function or approximate integration. For simplicity of computation and aided by a computer, let's use the Midpoint Rule and state:
$$T\approx\Delta t\sum_{k=0}^{n-1}\left(C\left(\frac{t_{k}+t_{k+1}}{2} \right) \right)$$
Now, we find that:
$$\Delta t=\frac{24}{n}$$
$$t_k=\frac{24k}{n}$$
$$\frac{t_{k}+t_{k+1}}{2}=\frac{12}{n}(2k+1)$$
Hence:
$$T\approx T_n=\frac{600}{n} \sum_{k=0}^{n-1}\left(\exp\left(-\frac{\left(\dfrac{12}{n}(2k+1)-15 \right)^2}{20} \right) \right)$$
Now, at the site
Wolfram|Alpha: Computational Knowledge Engine I used the command:
sum of (600/n)exp(-((12/n)(2k+1)-15)^2/20) for k=0 to n-1
where I substituted powers of 10 for $n$ and obtained (to 3 decimal places):
[TABLE="class: grid, width: 200"]
[TR]
[TD]$n$[/TD]
[TD]$T_n$[/TD]
[/TR]
[TR]
[TD]10[/TD]
[TD]197.814[/TD]
[/TR]
[TR]
[TD]100[/TD]
[TD]197.729[/TD]
[/TR]
[TR]
[TD]1000[/TD]
[TD]197.728[/TD]
[/TR]
[TR]
[TD]10000[/TD]
[TD]197.728[/TD]
[/TR]
[/TABLE]
Thus, the choice given by c) is the closest.