The tank has a volume of [tex]\dfrac\pi3R^2H[/tex], where [tex]H=6\,\rm m[/tex] is its height and [tex]R=\dfrac d2=2\,\rm m[/tex] is its radius.
At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:
[tex]\dfrac26=\dfrac rh\implies r=\dfrac h3[/tex]
The volume of water in the tank at any given time is
[tex]V=\dfrac\pi3r^2h[/tex]
and can be expressed as a function of the water level alone:
[tex]V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3[/tex]
Implicity differentiating both sides with respect to time [tex]t[/tex] gives
[tex]\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}[/tex]
We're told the water level rises at a rate of [tex]\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}[/tex] at the time when the water level is [tex]h=2\,\mathrm m=200\,\mathrm{cm}[/tex], so the net change in the volume of water [tex]\dfrac{\mathrm dV}{\mathrm dt}[/tex] can be computed:
[tex]\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}[/tex]
The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:
[tex]\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})[/tex]
We're told the water is leaking out at a rate of [tex]10,500\,\frac{\mathrm{cm}^3}{\rm min}[/tex], so we find the rate at which it's being pumped in to be
[tex]\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}[/tex]
[tex]\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}[/tex]
