Answer
${\bf 10^5}\;\rm oscillations $
Work Step by Step
We need to find how many oscillations are in the shortest-duration laser pulse that can travel through the fiber, and since the laser pulse is an electromagnetic wave packet, we can use the uncertainty principle for signal transmission
$$ \Delta t \cdot \Delta f \approx 1 \tag 1$$
where $ \Delta t $ is the pulse duration, and $ \Delta f $ is the bandwidth.
The number of oscillations in the pulse can be calculated by dividing the pulse duration by the period of one oscillation.
Recalling that the frequency $ f $ of the laser pulse is given by
$$ f = \frac{c}{\lambda} $$
where $f=\dfrac{1}{T}$ where $T$ is the period of one oscillation.
So,
$$ \dfrac{1}{T} = \frac{c}{\lambda} $$
Hence;
$$T=\dfrac{\lambda}{c} \tag 2 $$
We can find the number of oscillations $N$ by dividing the pulse duration $ \Delta t $ by the period $ T $.
$$ N= \frac{\Delta t}{T} $$
Plug $\Delta t$ from (1);
$$ N= \dfrac{\dfrac{1}{\Delta f} }{T} $$
Plug $T$ from (2)
$$ N= \dfrac{\dfrac{1}{\Delta f} }{\dfrac{\lambda}{c}} $$
$$ N= \dfrac{c }{\lambda \Delta f} $$
Plug the known;
$$ N= \dfrac{(3\times 10^8)}{(1.5\times 10^{-6})(2\times 10^9)} =\color{red}{\bf 10^5}\;\rm oscillations $$
