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I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $frac{ft^3}{hr}$, at what height above the drain will the elevation remain constant?

My attempt:

We are given that $$frac{dV}{dt}=kh$$ where $frac{dV}{dt}$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.

If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $frac{dV}{dt}=100frac{ft^3}{hr}$.

To find $k$, I must do separation of variables and integrate both sides, so: $$int{dV}=int{khdt}$$ $$V=kht+C$$
I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.

Solving for $k$: $$2500=k(5)(6)+0$$ $$k=frac{250}{3}$$
And solving for $h$ when $frac{dV}{dt}=100frac{ft^3}{hr}$:$$100=frac{250}{3}h$$ $$h=1.2ft$$
However, the answer is actually about $1.73ft$. Where is my mistake?

Your mistake is in the solution of differential equation and in the initial boundary condition.

see the following solution:
$$frac{dV}{dt}=kh$$
$$Afrac{dh}{dt}=kh$$
$$500frac{dh}{h}=kdt$$
integrate
$$500log h=kt+C$$
at $t=0, h=10ft$

so $$C=1151.3$$
at $t=6 hr, h=5 ft$
$$k=-57.76$$
now use that
$$frac{dV}{dt}=-57.76h$$
$$-100=-57.76h$$
so
$$h=1.73 ft$$

We have a differential equation that defines the rate that water leaves the drain.

$frac{dx}{dt} = -ax\
x = C e^{-at}$

We know the initial conditions and after 6 hours to find our constants.

$x(0) = 10$

This gives us $C$

$x(6) = 5\
5 = 10e^{-6a}\
6a = ln 2\
a = frac{ln2}{6}$

What do we know about the dimensions of the tank?

$v = pi r^2 x\
pi r^2 = 500$

The flow out the drain equals the flow into the tank.

$frac{dv}{dt} = frac {dv}{dx}frac{dx}{dt}\
500 frac {dx}{dt} = 100\
500(frac{ln 2}{6})x = 100\
x = frac{6}{5ln 2}$