Determining the current from an infrared LED used as photodiode
The amount of current that is produced by a photodiode is related to the light power it receives in watts and the "yield factor" in converting watts to amps. All recognized photodiodes have a specification for this. 

So, you begin by analysing the emitter: if an emitter is producing "so many" milli watts from (ostensibly) a point source, you can calculate the power density (watts per sq metre) at any distance. 

For an isotropic emitter, light power is emitted in all directions so, the power density at a given distance (R) is related to the surface area of a sphere (\$4\pi R^2\$) and the power originating from the "point source".

Given that a recognized photodiode will have a specified active surface area you can converts watts per sq metre back to incident power (watts). This then converts to amps via the "yield factor".

So you need to know how much light power is produced by the LED and how this power is concentrated in a particular direction. For instance, it is very unlikely that it will be isotropic and much more likely you can assume that at least 50% of the power produced is in a 3D angle of maybe 60 degrees. The LED spec should tell you that.

This "concentration factor" will give you watts per square metre at the distance your receiver is placed from the emitter. The active surface area of your receiver converts watts per sq metre to watts and then on to amps and you are done. Of course, the glucose solution will absorb some power so you also need to factor that in but, I guess that's the whole point of your experiment.

However, in your scenario you don't appear to know any of this so, unfortunately, you cannot know how much current to expect and, as others have said in comments, your only option is to try it - find an op-amp with low bias currents and try it out.

I would also recommend using a light source that is pulsed because then you can cancel out drift by sampling before light is emitted and then sampling whilst light is being emitted. This also allows you to get rid of the effects of ambient light changing the reading.