It is not clear in which part of the problem statement you experience issues.
However, a few hints may help you get the gist of such problems
- The sampling frequency, it is intuitive that in order for a system to monitor the state of another system it must poll it at higher rates than the highest rate in which the monitored signal is able to change its state. But what is the minimum sampling frequency that gives our sampling system the ability to correctly approximate the continuous signal in the discrete time domain?
This frequency threshold is at two times the highest frequency of the continuous signal:
$$
f_{sampling} \gt 2\cdot f_{bandwidth}
$$
The theorem that the problem statement suggests is the Shannon-Nyquist sampling theorem, the theorem that the above inequality comes from.
In real sampling systems, the sampling frequency is never taken exactly on this threshold, usually the threshold is taken a little bit higher than $2\cdot f_{bandwidth}$, for example at $2.2\cdot f_{bandwidth}$.
However, in theory it is safe to assume a sampling frequency at the Nyquist rate (i.e. the aforementioned frequency threshold).
Your analogue signal's highest frequency is $14.5\, Hz$, ranging from very low frequencies so it is safe to assume, this actually the bandwidth frequency.
All the above summarized in the following equation:
$$ f_{sampling} = 2\cdot f_{bandwidth}$$
- Bitrate, in order to calculate the bitrate of your sampling system it is necessary to know the sampling frequency. Fortunately, this was the first the step of the exercise. The second component for the bitrate calculation is the actual length of each sample in bits. The minimum number of bits that are necessary in order to represent a quantized value in x distinct levels is given by:
$$
2^{bitlength}\, =\, x
$$
or
$$
bitlength = \lceil\log_2{x}\rceil
$$
You could actually deduce the bitrate equation by performing a dimensional analysis, because it is stated that it is measured at kbps(kilobits per second). For instance, we know that $Hz$ is measured in $\frac{samples}{sec}$ and kb stands for kilobits meaning $1000\, \frac{bits}{sample}$, so:
$$
bitrate = bitlength\cdot f_{sampling} \left[\dfrac{bits}{sample}\dfrac{samples}{sec} \right]
$$
or in kbps
$$
bitrate = bitlength\cdot f_{sampling}\cdot 10^{-3} \left[kbps\right]
$$