How to Compute bias Resistor for phototransistor Value?
100 nA is the dark current. That is, with no light falling on the detector, it will put out about 100 nA.

0.8 volts is the saturation voltage. That is, if you shine a light on the transistor when it is connected like 







simulate this circuit &ndash; Schematic created using CircuitLab

when the light is dim, VOUT will be nearly Vcc. As the intensity increases, VOUT will drop, but it will never get lower than 0.8 volts.

Now you need to estimate what light level you expect to detect. Let's say you want the transistor to reach saturation with a power level of 1 mw/$cm^2$. From the first table you can typically expect 0.2 mA, but it might be 0.1 mA. So $$R = \frac{V}{i} = \frac{5-0.8}{.0001} = 42k$$ as your calculation showed. However, this means your pot should be either 50k or 100k, depending on what's available. If you use a 10k pot, the most voltage across the resistor you can hope for is $$V = iR = .0001 \times 10,000 = 1 volt$$ This would mean that, for 1 mW/$cm^2$ illumination, your output voltage would not get below 4 volts.

For greater sensitivity you need more resistance, not less.