How do you find the resolution and sampling rate needed for a certain signal to noise ratio of an analog to digital converter?
SNR required for your ADC is \$10log_{10}(50000) = 47dB\$. This will be contributed by your thermal and quantization noise. To keep some scope for the thermal noise we need quantization noise better than 47dB. For instance, if you choose signal to quantization noise ratio (SQNR) of 50dB, then the required resolution becomes:
$$SQNR = 6.02*N + 1.76 \implies N = 9\ bits$$
The thermal noise is decided by the size of capacitors you choose and is given by:
$$NF\frac{kT}{C},$$
where, k = Boltzmann Constant, NF = noise factor for your system, which includes the contribution of noise from your op-amp and sampling noise etc. It is design dependent factor.
Based on this and your thermal noise margin, you can calculate the size of the capacitors required. But usually at such low resolutions designs are not thermal noise limited.
The bandwidth of the ADC gives you the upper bound on the sampling rate you can achieve, and if you are going for a Nyquist Rate converter the maximum input frequency which you can apply without aliasing would be half of this rate.
The bandwidth, in turn, also depends on the size of the capacitors used, higher the capacitors lower the bandwidth. So, there would be a trade-off between bandwidth and resolution. Higher resolution would require you to have higher capacitors to decrease noise but your bandwidth also goes down, so you have to make some choice based on your requirements.