Finding steady state concentration given half life and production rate
tl;dr besides making this a bit harder than necessary, your main problem is that you have confused the lifetime (average time to removal of a molecule), not the half-life (time until half of the molecules are removed, which is $\ln(2)$ times the lifetime ...) Wikipedia has some formulas ...

You can translate the problem into a differential equation (not as scary as it sounds):

$$
\frac{dB}{dt} = \underbrace{1000\vphantom{\frac{1}{200}}}_{\textrm{production}} - \underbrace{\frac{1}{200} B}_{\textrm{removal}}
$$

The only tricky part here is recognizing that if the lifetime is 200 s, a fraction 1/200 of the existing molecules will be removed per second. 

Now we have to solve this at equilibrium: $dB/dt =0$. We get
$$
\begin{split}
1000 - \frac{B}{200} &amp; = 0 \\
1000 &amp; = \frac{B}{200} \\
B = 200000
\end{split}
$$

Or you could memorize Little's Law: steady state = arrival (1000/s) $\times$ lifetime (200 s).