Answer
${\bf 10^5}\;\rm pulse/s$
Work Step by Step
We know that
$$\Delta f \approx \frac{1}{\Delta t}\tag 1$$
The period of the pulse train is given by
$$ T=2\Delta t $$
Plug $\Delta t$ from (1);
$$ T= \frac{2}{\Delta f}\tag 2 $$
where we know that
$$f=\dfrac{1}{T}$$
So, plugging from (2)
$$f= \frac{\Delta f}{2}$$
where $f$ here is the maximum transmission rate.
Plugging the known;
$$f= \frac{(200\times 10^3)}{2}=\color{red}{\bf 10^5}\;\rm pulse/s$$
