Consumer Batteries Needed to Ignite a 10ohm Resistor
Ohm's law says sqrt(P/R) = I. So your resistor can handle at least sqrt(0.25/100) = 50mA (not much). 

In practice, because parts are designed to do exactly the opposite of what you want (e.g. not explode), they can handle much more than that (don't ever design in parts beyond spec in products or anything you ever share with anyone!!!).

10X that (>500mA) should do the trick, but that doesn't necessarily guarantee that it's internal ignition will result in external flame. Many parts are coated to prevented this and the chemistry is always being improved to further reduce the risk of a secondary ignition. That's how products catch fire (a VERY bad thing).

To get 500mA through 100 Ohms you need V = IR = 0.5 * 100 = 50V source. That's about 30 E91 ("AA") batteries in series.

If you reduce your resistor to an 1/8W 1 Ohm part, you can relax your battery requirements much more. Repeating the math...

sqrt(0.125/1)*1 = 350mV, something easily achieved in just a single "AA" battery. At 1 Ohm, the current demand will be 1.5A (approx) from an "AA" battery, so you need a battery with 1C rating > 2A for safety (so that the battery doesn't ignite too!).

I don't think this is necessarily a bad question. Since it's clear that the OP is doing something dangerous that others shouldn't attempt. Sometimes you have to experiment "on the edge" to truly grasp concepts... just sayin'