Simple IR pulse detector with IR phototransistor
The voltage accross the resistor is the current thru it times its resistance.  That's basic Ohm's law.  The values you give are a bit conflicting and misleading, but let's say the maximum light current is 20 mA.  If the power supply is 5 V and you want 500 mV on the transistor so it doesn't saturate, then that leaves 4.5 V on the resistor.  4.5V/20mA = 225 &Omega;.

However, things are not that simple.  The ambient light level can vary by a large margin, so the DC dynamic range has to be high.  To be safe, you make sure that the resulting voltage stays linear with light up to the maximum light the transistor is rated for or you know will ever be encountered.  That's what I did in the example above.  The problem with that is that gain will be very low.  The signal you want may only be a millivolt or possibly less.

This is why light sensors like this are almost always AC coupled.  Use the largest resistor that still allows about 500 mV accross the transistor with maximum light, then AC couple that into a amplifier.  The ambient light level is DC or slowly changing, so will get blocked.  Now you can amplify the AC signal in the frequency range you are looking for.  This may require a voltage gain of 100s to 1000 or more.  Anything you can do mechanically to eliminate ambient light pickup and strengthen the intended received beam will help.  With lower maximum ambient light, you can make the resistor larger, which increases gain.