The radius of the water surface is 1/3 the height of the water.
let the height be h and the radius be h/3 (in cm)
$$\\V=\frac{1}{3}\pi r^2 h\\\\
V=\frac{1}{3}\pi(\frac{h}{3})^2 h\\\\
V=\frac{\pi h^3}{27} \\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}$$
$$\\When \;\;h=200\qquad radius=200/3\\\\
V=\frac{\pi\times 200^3}{27}=\frac{8\times 10^6 \pi}{27}\\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}=\frac{\pi\times 200^2}{9}=\frac{4\times 10^4 \pi}{9}\\\\
\frac{dh}{dt}=20\\\\\\$$
$$\\\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}\\\\
\frac{dV}{dt}=\frac{4\times 10^4 \pi}{9}\times 20\\\\
\frac{dV}{dt}=\frac{80\times 10^4 \pi}{9}\\\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-\frac{dV_{out}}{dt}\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-11000\\\\
\frac{dV_{in}}{dt}=\frac{80\times 10^4 \pi}{9}+11000\\\\
\frac{dV_{in}}{dt}=290,253\;\;cm^3/min\\\\$$
I think the water is being pumped into the tank at the rate of 290 litres per minute to the nearest litre. (edited)
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LancelotLink will probably have some great retort for my benefit LOL
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Okay LancelotLink has has some fun at my expense. He got Chimp Ayumu to do it. She is one smart Chimp.
Our answers are still different. Mine does look rather huge. 
I'll have to look some more!
How annoying. 
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It's all fixed now and I have a very good excuse!
Someone changed the metric system. Who did that!
Now that there are only 100cm in a metre instead of my original 1000cm in a metre, it works perfectly.
DOH!
