As depicted the 50 Hz and 60 Hz rejection filter is not necessarily a CIC filter (but could be constructed with one when the sampling rate is an integer multiple of $50 \times 60$ such that 50 and 60 are factors. However it is much more likely that they would be constructed as the cascade of two 2nd order IIR nulling filters as detailed in this post, and summarized below:

This is a "2nd Order Biquad" filter, and the frequency response for this structure with the coefficients as given above is:
$$ H(z) = K\frac{z^2-2z\cos\omega_n+1}{(z^2-2az\cos\omega_n+a^2)} $$
Where $K$ is a normalizing gain constant, if desired for adjusting the filter gain in the passband regions to be 1, given as:
$$K = \frac{1+2a\cos(\omega_n)+a^2}{2+2\cos(\omega_n)}$$
$\omega_n$ is the notch frequency in normalized radian frequency in units of radians/sample, determined by dividing the radian frequency by the sampling rate.
$\alpha$ is the radius to the pole for the filter, which for narrow bandwidths is close to 1. The value for $\alpha$ is approximately $1-\omega/2$, where $\omega$ is the 3 dB bandwidth of the notch in radians/sample. The exact solution is detailed in this post. For high-Q designs (tight notch) and for relatively high sampling rates compared to the notch frequencies the 2nd order biquad (as this is called) can be challenging to implement with fixed-point designs, and it is imperative to confirm results with bit accurate simulations. If available precision is an issue, alternate structures exist as suggested in this blog post by Rick Lyons, including the "Coupled Form" by Rader and Gold (1967) and improved structures by Agarwal and Burrus (1975).
For example if the sampling rate was 1200 Hz, the first filter associated with 50 Hz would have:
$$\omega_1 = \frac{(2\pi) 50}{1200}$$
And the second filter (repeating the structure above with the two filters in series) with the notch at 60 Hz would have:
$$\omega_2 = \frac{(2\pi) 60}{1200}$$
If we wanted a 1 Hz bandwidth notch, then using the approximation:
$$\alpha = 1 - \frac{(2\pi) 5}{1200} = 0.99476$$
With the resulting frequency response:

fM/Nis the notch frequency... but as response figure show, the notch is not an simple integer multiple of 50Hz nor 60Hz. And if realized in this way, the settling time is not guaranteed...Could you further explain these issues? – Ring Feb 13 '23 at 11:56