Typical Calculus BC Separation of Variables Question

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

see the following solution: $$\frac{dV}{dt}=kh$$ $$A\frac{dh}{dt}=kh$$ $$500\frac{dh}{h}=kdt$$ integrate $$500\log h=kt+C$$ 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 $$h=1.73 ft$$