Maximum sensitivity of a wheatstone bridge
To begin, I don't believe the question is worded properly. A moment around y will not affect B and D differently. I have a feeling they meant to say B and D are used for axial load and A and C are used for measuring the moment.

A moment measured at two sides of that object would create equal and opposite forces. In a strain gage, it would come back as equal and opposite changes in resistance.
A = Rg + x
C = Rg - x  

A Wein bridge is nothing more than two voltage dividers while measuring the difference between them. A voltage divider has the algebraic form:  



$$V_{out}=V_{dd}\cdot \frac{R_{bottom}}{R_{bottom}+R_{top}}$$

To determine the sensitivity of this setup, you could simply look at the max/mins of it during different conditions. At a maximum we could say that x would be the same size as R. Otherwise you could get negative resistances. With that information and the 4 unique topologies that present themselves let's determine what voltage each vertical path would give as x varies from -R to R. Let's assume Vdd is 1 for simplicity/normalization.

Parallel on top:
A C
R R
If A goes up, C goes down: 1/3 &amp; 1. Vdiff = -2/3
If A goes down, C goes up: 1 &amp; 1/3. Vdiff = 2/3
Total range = 4/3

Parallel on bottom:
R R
A C
If A goes up, C goes down: 2/3 &amp; 0. Vdiff = 2/3
If A goes down, C goes up: 0 &amp; 2/3. Vdiff = -2/3
Total range = 4/3

Serial:
A R
C R
If A goes up, C goes down: 0 &amp; 1/2. Vdiff = -1/2
If A goes down, C goes up: 1 &amp; 1/2. Vdiff = 1/2
Total range = 1

Diagnol:
A R
R C
If A goes up, C goes down: 1/3 &amp; 0. Vdiff = 1/3
If A goes down, C goes up: 1 &amp; 2/3. Vdiff = 1/3
Total range = 2/3

As we can see just from this simple intuitive analysis, placing the sensors parallel in the Wein bridge gives the greatest range whether they're at the top or the bottom.



The lengthy and more mathematically robust way would be to use the full Vdiff equations for each topology and take the derivative with respect to x to determine the sensitivity. 

I did one topology (bottom parallel) for example and ended up with this:


$$
V=\frac{R}{2R+x}-\frac{R}{2R-x}=\frac{R(2R-x)-R(2R+x)}{(2R+x)(2R-x)}=\frac{-Rx-Rx}{4R^2-x^2}=\frac{2Rx}{x^2-4R^2}
$$



$$
\frac{dV}{dx}=\frac{2R(x^2 - 4R^2) - 2Rx(2x)}{(x^2-4R^2)^2}=\frac{-2Rx^2-8R^3}{(x^2-4R^2)^2}
$$

If you also calculate the serial case which is very simple you'll get dV/dx for that case being -1/(2R). And if we plot these against each other with R normalized to 1 we see that around 0 they have the same sensitivity, but as you go further away from 0, the parallel case begins to get more sensitive.



Also looking at the original Vdiff plots will also show you that putting them in parallel will increase the sensitivity further away from x=0:


Lastly, it should be noted that although some of the solutions increase the sensitivity further away from x=0, they do at the cost of linearity. The problem says maximum sensitivity, so the parallel case is the way to go. In actual engineering though, it may be wisest to use the serial case because it gives a linear sensitivity allowing a simple conversion between a voltage and a strain or moment.