How much power do I need to transmit a radio signal through a solid ice crust 100 km thick?
I can't answer that directly but Nasa are probing Greenland's ice sheets with aeroplane radar to find the depth of the bedrock. Here's what they say about ice and radio waves: -


  Ice, on the other hand reacts differently depending on the radar's
  frequency. It reflects high-frequency radio waves, but despite being
  solid, lower frequency radar can pass through ice to some degree. This
  is why MCoRDS uses a relatively low frequency—between 120 and 240 MHz.
  This allows the instrument to detect the ice surface, internal layers
  of the ice and the bedrock below. "To sound the bottom of ice you have
  to use a lower frequency," said John Paden, CReSIS scientist. "Too
  high a frequency and signal will be lost in the ice."


This came from here and it's interesting to note that this is radar and requires a reflection from the bedrock to pass back through the ice to the receiving aeroplane. I would imagine that the reflected power is a fraction of the incident power reaching the rock so maybe you might get 10x this distance thru a solid ice sheet with a one-way transmission.

Here's the sort of image they are getting: -



It looks to me that +3km is possible with radar. I don't know what the radar beam angle is so it's impossible to calculate what the incident power at the surface of the ice is - the transmission from the aeroplane might be a 1MW pulsed radar with a very tight beam angle producing an incident power at the top surface of the ice of hundreds of watts. Also, the reflection from the bedrock will not be a tight beam - this means the power reflected back will be spread thinly as distance increases (see Friis equations). Also the power received at the aeroplane will be much smaller than that emitting from the surface of the ice - again see the Friis equations.

Addendum

I had a think about the link loss for the radar application: -


Link-loss from plane to ground. A 2m diameter dish will have a gain of \$\frac{\pi^2 D^2}{\lambda^2}\cdot 0.6\$ = 3.35 or about 10.5 dB. If the aircraft is 1km above the ice, the link loss to an identical dish (on the ice) will be -21 (antenna gains) +32.5 + 20log(MHz) + 20log(km) = 11.5 + 46 + 20 = 78dB link loss. If the radar output power is 1 MW (90 dBm), the received signal at the ground/icecap surface will be 12 dBm (16 mW).
It's the same problem for the reflection signal. At the surface, it is subject to the same attenuation up to the aircraft (78dB) that is 1km higher.


These losses won't be encountered by a simple transmission thru ice - transmitting and receiving antennas are sited either in the ice or at its surface. This all bodes well for being able to transmit in a single direction through large distances of ice.