Bad news. A bleeder is NOT recommended for your (10mA) higher leakage current sensor. To research this, go to:
http://www.ab.com/manuals/sn/871-2_0.pdf
and see page 105.
Note that the maximum leakage current listed is 6.5mA which calls for a 4.7K ohm, 3 watt resistor. To compensate for your higher leakage current, the bleeder would have to be of lower resistance. BUT CAUTION! when the sensor turns on, the full voltage across the resistor would generate enough wattage to get VERY HOT. This is one reason why bleeder resistors are not generally recommended for higher leakage current sensors like the one you are trying to use. (And why your 10mA monster is "off the charts"). You SHOULD consider buying a different sensor for your application.
Or maybe using an interposing relay might help - but be sure it's not a sensitive one or the leakage current will keep it "on" even when it's supposed to be "off" and you'll be right back to square one.
Also, if you're just experimenting (and not working on an industrial application) you might try this other "trick". Let the sensor turn on a full-voltage incandescent lamp (not an LED or neon lamp) and connect your SLC input in parallel with the lamp. In essence, the lamp becomes the bleeder resistor and of course it's already designed to dissipate the heat created by the full wattage of a "sensor on" condition. You'll probably need at least a 7 watt bulb - experiment until you find one that works. But be aware that when the lamp bulb eventually burns out, the SLC will always see the input as "on" even when the sensor is "off". Whether this is safe or not depends on your particular application.
Also, you posted your original question on the "practice" board. That's why you didn't get many answers. You should put your future posts on the "Live Question and Answer" board so that more people will see them and be able to help.
Best of luck.