Your math for the resistor divider is correct, but the output voltage will change it you connect anything to it. You can't draw 100mA from it. The output will sag to zero at a 16mA load.
Working Voltage: 3.0V +/- .1V
That's 3%, that requires already precision components. Standard components are often 5%. Also keep in mind that the 3.3V may have a 5% tolerance as well!
A series resistor was suggested. That's the cheapest solution. And the worst. It's a Bad Habit™ you never should adopt. The voltage drop across the resistor changes with the load, so that you don't get a properly regulated 3V.
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I'm getting some critique on this. The reason I'm against a series resistor as a voltage regulator is that basic specs for a regulator, like line regulation and load regulation are very bad. But! This seems to be a special case. The load is a Wheatstone bridge which has a rather constant resistance, probably far more constant than the 10% the spec mentions (which is probably process tolerance, not measurement variation). A constant load means a constant voltage drop across the resistor, so that fixes the load regulation question. Then line regulation. Input is 3.3V from an iPhone. The voltage suggests this is regulated too, but we don't know for sure. If it's regulated then there's no line regulation issue either, and the series resistor can be seen as half of a resistor voltage divider, dividing a fixed voltage. In that case a series resistor will work. I have no problem admitting that. I still have to see the 3.3V confirmed, though. If it can vary the resistor is still bad, regardless of the constant load! In most cases either the input voltage or the load (or both) is variable, and then a series resistor won't give you a properly regulated, fixed output voltage.
Figures: the series resistor will give you a line regulation of 910mV/V. Compare with an LM7805, which has a typical line regulation of 0.2mV/V.
The series Schottky diode's voltage also varies a bit with current, but there's also process variation. Tony's graphs show typical and maximum voltage drops, but if you want to use one it's the minimum voltage drop which is important as well. Check it before use. Especially if the 3.3V would have a +5% tolerance (=3.47V). Tony's diode only drops 250mV typical, at this input voltage that would be too little. This diode only drops 150mV at 100mA, so that wouldn't even do at the nominal input voltage.
The zener diode that Kaz proposed also suffers from process variations. At 5mA this 3V zener can have a reverse voltage between 2.85V and 3.15V, which is outside the limits you specify. Most zener diodes will have this kind of variation (5%). Precision zeners (2%, 1%) exist, though. Your zener will have to draw at least 10mA, preferably 15mA (5 to keep it stabilizing, 10 to cover the 10mA tolerance on the load), which may or not may be within your budget.
A zener with a differential resistance of 30\$\Omega\$ drawing 15mA has a line regulation of 860mV/V, so that's only as good (or bad) as the resistor.
The neatest solution is an LDO regulator. It's a bit more expensive but the specs are worth it. My favorite for low power applications is the Seiko S-812C (only 1\$\mu\$A ground current!), but that can't supply the required 100mA. The Torex XC6210 however can, even up to 700mA. 2% accurate, only 50mV dropout voltage and only 35\$\mu\$A ground current.
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Typical application for the XC6210:
Don't forget the capacitors, especially the output cap. And don't forget to connect the Chip Enable (CE) to \$V_{IN}\$.
I know the answer is subjective.
I use this whenever there is a need to switch between two supplies with at least 1 V difference automatically. Here, assuming VIN_EXT is less than 5 V, the circuit will draw power from PC whenever USB power is available, else from VIN_EXT automatically.
simulate this circuit – Schematic created using CircuitLab
Major advantage here is that, drop across the FET will be just a few 10s of mV at max compared to diode drop of about 300 or 400 mV
Best Answer
Although the datasheet doesn't explicitly state a minimum load current, it does give the output regulation rating for 1 mA to 40 or 100 mA, implying to me that there is a minimum load required for the part to regulate (and I recall the older 78xx regulators specifiying a minimum 5 mA load current for proper operation).
Put a 1 mA or more load on it and see if it works.