Answer
$\approx 236\;\rm oscillations$
Work Step by Step
To find the number of oscillations when the amplitude is 30% of the initial amplitude $0.3A$, we need to find the period of one oscillation, the time constant, and the time it takes to reach this small amplitude.
We know, for a damping oscillation, that
$$A_{(t)}=Ae^{-t/2\tau}$$
At $t=50$ s, the amplitude is $0.6A$, so
$$0.6\color{red}{\bf\not} A=\color{red}{\bf\not} Ae^{-50/2\tau}$$
So that
$$\dfrac{-50}{2\tau}=\ln(0.6)$$
And hence the time constant is
$$ \tau=\dfrac{-50}{2\ln(0.6)}=\bf 48.94\;\rm s\tag 1$$
The time it takes to reach $0.3A$, $t_1$
$$0.3\color{red}{\bf\not} A=\color{red}{\bf\not} Ae^{-t_1/2 (48.94)}$$
Hence,
$$\dfrac{-t_1}{2(48.94)}=\ln(0.3)$$
$$t_1=-2(48.94)\ln(0.3)=\bf 117.845\;\rm s\tag 2$$
The number of oscillations is given by
$$N=\dfrac{t_1}{T}$$
where $T=1/f$, so
$$N=t_1f$$
Plugging from (2) and plugging the known;
$$N=(117.845)(2)=235.69\;\rm oscillations $$
Therefore,
$$N\approx\color{red}{\bf 236}\;\rm oscillations$$
