Potential energy = mass × g × height. So the rider would gain 100 × 9.81 × 100 = 98100 joules on the climb.

With a 10% slope, the 100 m tall hill would have a run of 1000 m. But the slope length would be fractionally longer according to Pythagorean’s theorem. The distance ridden would therefore be the square root of (100^2 + 1000^2) or 1005 m. Okay, not much difference.

10 km/h is 2.77 m/s, so covering the 1005 m would take 362 s.

A watt is defined as a joule per second (SI units are beautiful), so expending 98100 J in 362 s would require 271 W. That’s the power needed to drag the weight up the hill against gravity at that speed.

But rolling a bicycle on flat ground at 10 km/h takes in the region of 20 W according to

Bike Calculator, so the total power for climbing the hill at 10 km/h on a still day would be very close to 290 W. That’s likely more than you can sustain for six minutes.

A food calorie is about 4.2 kJ. As calculated above, this climb would require in the region of (362 s × 290 W) = 105000 J, which is therefore about 25 calories. However, humans are only about 25% efficient at converting food energy to work done, so you’d need to eat about four times that or 100 calories to replace the energy used on the climb.

EDIT: in short, I agree with sib. Maybe my working out was useful.