Can we choose a sampling frequency to remove unwanted noise at a specific frequency?
Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at $\pi/2$.

So what you could do is use the relationship between the angular frequency $\omega$ in [radians/sample] and the relative frequency $f$ in [cycles/sample]
$$
\omega = 2\pi f = 2\pi \left(\frac FF_s\right)
$$
Then compute $F_s$ for $F = 80 \ \rm Hz$ and $\omega = \pi/2$. You then get
\begin{align}
\frac \pi 2 &amp;= 2\pi \left(\frac{80}{F_s}\right)\\
\implies \frac 12 &amp;= 2\cdot \left(\frac{80}{F_s}\right)  \\
\iff 80 &amp;= \frac 12\cdot \left(\frac{F_s}{2}\right)
\end{align}
Giving you the $320\ \rm Hz$.
In MATLAB
You can plot the magnitude response from the frequency response using MATLAB's freqz function where you can specify the frequency grid from $0$ to Nyquist (i.e. $\pi$). With your resulting $320\ \rm Hz$, you can run the function with this $F_s$ and see that $80\ \rm Hz$ is indeed suppressed.