A question about instrumentation
As has been mentioned in some of the comments, your overall noise floor of 20 mV RMS is far larger than your quantization noise, so without doing any calculations, you can be sure that halving your dynamic range while using the same DAQ device will have negligible impact on your SNR. However, you requested a quantitative analysis, so I will start by quantifying the ratio of the transducer noise power to the quantization noise power.

The quantization error is modeled as being equally likely to take on any value within a range of one least significant bit. This is a good assumption as long as your signal is much larger than your least significant bit. With this assumption, one arrives at the well-known result for the time-averaged quantization noise power \$P_{quant} = \frac{LSb^2}{12}\$, where \$LSb\$ is the voltage associated with the least significant bit (reference here). In this case, the LSb value is \$LSb = \frac{20 mV}{2^{24}}\ = 1.2 \mu V\$. Therefore 

$$P_{quant} = \frac{(1.2 \mu V)^2}{12} = 0.12 (\mu V)^2$$

Your transducer noise power in the first case (full scale  \$\pm 10V\$ output, rms noise voltage 20 mV) is \$(20 mV)^2\ = 400\cdot 10^6(\mu V)^2\$. So your transducer noise power \$P_{tn}\$ is larger than your quantization noise by

$$
10log(\frac{P_{tn}}{P_{quant}})=10log(\frac{400\cdot 10^6 (\mu V)^2}{0.12 (\mu V)^2})  = 95 dB
$$

If you halve the output of your transducer signal so that the noise is now 10 mV rms, you lose just 6 dB from this ratio so that your transducer noise is 89 dB larger than your quantization noise. Since your transducer noise overwhelms your quantization noise to such a high degree, at this point I should stop writing, or you should stop reading and be done. But I just had some coffee, so I'll carry it through and calculate the SNR for a full-scale sine wave output from your transducer.

SNR

A full-scale sine wave would be a sine wave of peak-to-peak amplitude of 20 V in your first scenario, and 10 V in your second scenario. Using this full-scale value leads to the maximum SNR you could have for a sine-wave signal.

The SNR is the ratio of the signal power to the noise power. For a pure sine wave of peak-to-peak amplitude \$V_{pp}\$, the signal power \$P_{sig}\$ is simply the time average of \$V(t)^2\$, which is 

$$P_{sig} = \frac{V_{pp}^2}{8}$$

The total noise power \$P_{Noise}\$ is the sum of the transducer noise power and the quantization noise power, or

$$P_{Noise} = P_{tn}+P_{quant}$$

and from that you can calculate your signal-to-noise ratio

$$
SNR = 10log(\frac{P_{sig}}{P_{tn}+P_{quant}})
$$

Let's evaluate this expression to ridiculous precision in each case just to illustrate the negligible effect the quantization noise will have on the SNR. The quantization noise power \$P_{quant}\$ will be the same \$0.12 (\mu V)^2\$ in both cases.

Case 1: \$V_{pp} = 20 V\$, \$V_{noiseRMS} = 20 mV\$

\$P_{sig} = (20 V)^2/8 = 50 V^2\$

\$P_{tn} = (20 mV)^2 = 400\cdot 10^{-6} V^2\$

\$SNR = 50.9691001287 \$ dB

Case 2: \$V_{pp} = 10 V\$, \$V_{noiseRMS} = 10 mV\$

\$P_{sig} = (10 V)^2/8 = 12.5 V^2\$

\$P_{tn} = (10 mV)^2 = 100\cdot 10^{-6} V^2\$

\$SNR = 50.9691001248 \$ dB

The SNR has been kept to absurd precision so you can see just how far out we need to go to see the difference. You can see that the quantization noise is completely and utterly negligible here, and your focus should be on reducing the transducer noise through filtering and averaging.

This is the maximum possible SNR of your system for a full-scale sine wave transducer output. If your output signal is only 1/10th the full-scale voltage, you would subtract 20 dB from your SNR.

As a side note, I will make the same point already made by Paul Uszak in his comment that your DAQ is unlikely valid all the way to 24 bits, and so this 89 dB ratio of transducer noise to quantization noise is actually a high estimate. But 89 dB still gives you so much headroom, you can remove many bits of real resolution and still have your quantization noise absolutely swamped by your transducer noise.

So you asked if one scheme was better than the other. I think the answer is that you are free to take the quantization noise out of that decision, and decide based on other system considerations.