To follow up:
[MATH]x\frac{\text{hL}}{\text{min}}=x\frac{\text{hL}}{\text{min}}\cdot\frac{100\text{ L}}{1\text{ hL}}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{10}x\frac{\text{m}^3}{\text{min}}[/MATH]
And so:
[MATH]0.75\frac{\text{hL}}{\text{min}}=0.075\frac{\text{m}^3}{\text{min}}[/MATH]
Thus, the time \(t\) required to fill the tank of volume \(V\) at the rate \(r\):
[MATH]t=\frac{V}{r}=\frac{10.2\text{ m}^3}{0.075\dfrac{\text{m}^3}{\text{min}}}=136\text{ min}=2\text{ hr}\,16\text{ min}[/MATH]