The higher the water level in the tank, the faster the flow will exit from the orifice in the bottom. How fast will it exit? Viscous effects are of course present in the flow, but they can be neglected for the streamtube which exits the container, so we can employ Bernoulli's equation. Mathematically, the flow variables will conform to the following relation:
$P+\frac{1}{2}\rho V^2+\rho gz=P_0=Const.$
Where $P$ is the fluid static pressure, $\rho$ is the fluid density ($\rho=1000kg/m^3$ in this case), $V$ is the local velocity, $g$ is the acceleration of gravity $(g=9.81m/s^2)$, $z$ is the height above the reference plane, and $P_0$ is the stagnation pressure. Now, the key insights for this problem are twofold. The first is to realize that the flow pressure is equal to atmospheric both at the top of the body of water and at the bottom $(P_{top}=P_{bottom}=P_{atm})$. Additionally, the velocity of the flow at the top is all but negligible in comparison to that of the jet flowing out of the orifice $(V_{top}\cong0)$. Bernoulli's equation, with these assumptions applied, reduces to:
$\rho gh=\frac{1}{2}\rho V_{jet}^2$ ($z=0$ is the bottom of the container).
Because the density of water is essentially constant, we can simply divide it out and rearrange our equation to leave us with:
$\boxed{V_{jet}=\sqrt{2gh}}$.
This result could have been derived purely from the principle of Conservation of Energy, but it is instructive to rigorously derive it in fluid dynamic terms. Thus we can see that the higher the water level in the tank, the faster will be the velocity of the water jet, and (for a given cross-sectional area), the flowrate will increase proportionately $(Q=VA)$. There is only one height for which the flowrate entering the container will be equal to the flowrate exiting, I trust that you can solve the problem from here.
