The normalized low pass Butterworth filter:
The normalized low pass Butterworth filter is a type of filter that has a flat frequency response in the passband and rolls off gradually in the stopband. It is characterized by its order, which determines the sharpness of the roll-off.

Designing an analog bandpass filter:
To design an analog bandpass filter, we need to transform the normalized low pass Butterworth filter to the desired frequency range by applying frequency scaling and frequency transformation techniques.

Given specifications:
- Upper cutoff frequency: 20 kHz
- Lower cutoff frequency: 50 Hz
- Stopband attenuation: 20 dB
- Stopband frequencies: 20 Hz and 45 kHz

Frequency scaling:
To scale the normalized low pass Butterworth filter to the desired frequency range, we use the following formula:
f = f0 * (f2 - f1) / (f - f1)

Where:
- f is the frequency in the desired range
- f0 is the frequency in the normalized low pass Butterworth filter
- f1 and f2 are the lower and upper cutoff frequencies of the desired range

By applying the frequency scaling formula, we can determine the new cutoff frequencies for the bandpass filter:
- Upper cutoff frequency: f2 = 20 kHz
- Lower cutoff frequency: f1 = 50 Hz

Frequency transformation:
Once we have the new cutoff frequencies, we can perform a frequency transformation to obtain the transfer function of the bandpass filter. The frequency transformation formula is given by:
s = (s^2 + ω0^2) / (Bs)

Where:
- s is the Laplace variable
- ω0 is the geometric mean of the lower and upper cutoff frequencies
- B is the bandwidth of the bandpass filter

Stopband attenuation:
The stopband attenuation determines the sharpness of the roll-off in the stopband. A higher stopband attenuation corresponds to a steeper roll-off. In this case, the stopband attenuation is 20 dB.

Order determination:
The order of the normalized low pass Butterworth filter required to meet the given specifications can be determined using the following formula:
N = log10((10^(A/10) - 1) / (10^(B/10) - 1)) / (2 * log10(ωc))

Where:
- N is the order of the filter
- A is the stopband attenuation in dB
- B is the transition band attenuation in dB
- ωc is the cutoff frequency in radians per second

In this case, the stopband attenuation (A) is 20 dB and the transition band attenuation (B) is 3.0103 dB. By substituting these values into the formula, we can calculate the order of the filter.

Calculation:
N = log10((10^(20/10) - 1) / (10^(3.0103/10) - 1)) / (2 * log10(ωc))

By solving this equation, we find that the order of the filter is approximately 3.

Therefore, the correct answer is option 'B' (3).