Even without multiplying by $1000$, $A(50)$ turns out
to be $\approx 40\, km^2$, which is a lake more than a tank (?!).
If so, it will take approx. one month to lower the level by $1\,m$.
Also, $b(t)$ is becoming negative before reaching full opening in about $2.5\, s$
Your problem is concerning more hydraulic Engineering
than mathematics and you should move it [to this site]
(https://engineering.stackexchange.com/).
However, all the problem
consists in applying energy and mass conservation.
In a very first approximation (ideal case) you will have that
$$
\eqalign{
& \left\{ \matrix{
\rho \,A(t)\,dh = \rho \,A(t)\,\dot h(t)dt = \rho \,2\,b(t)\,v(t)dt \hfill \cr
\rho \,A(t)\,\dot h(t)dt\,h(t) = {1 \over 2}\rho \,2\,b(t)\,v(t)dt\;v(t)^{\,2} \hfill \cr} \right. \cr
& \left\{ \matrix{
\,A(t)\,\dot h(t) = \,2\,b(t)\,v(t) \hfill \cr
A(t)\,\dot h(t)\,h(t) = \,b(t)\,v(t)\;v(t)^{\,2} \hfill \cr} \right. \cr
& \left\{ \matrix{
\,v(t) = \sqrt {2\,\,h(t)} \; \hfill \cr
\,\dot h(t) = \,{{2\,b(t)} \over {A(t)}}\sqrt {2\,\,h(t)} \hfill \cr} \right. \cr}
$$
where $\rho$ is the density and $v$ the (mean) outgoing speed,
and where it is assumed that $h$ and thus $A(h)$, are enough high
that it is possible to assume that the kinetic energy of the top level be practically null.
For $h=50$ this would give $v=10\,m/s$ and a debit of max $20\,m^3/s$.
But in the reality that would be an unattainable max, since losses comes into play
which depend on the geometry of the outlet, rugosity of its walls, actual water speed profile
etc.. Correction coefficients shall be introduced accordingly, [see for instance this article]
(https://www.engineeringtoolbox.com/sluice-gate-flow-measurement-d_591.html).
