Earth's volume is 9.38e11 km^3. Or 9.38e23 liters (dm^3, cubic decimeters).

9.38 * 10^11 km^3 / (2^64) in m^3 = 50.8 m^3.

So you'd have one 64-bit address for every 50.8 cubic meters of earth.

Also, what's the volume of the solar system?

Based on the GP post, I believe the solar system is a sphere with radius 16.4 km. The volume would, therefore, be almost exactly 2^64 cubic centimeters.

Yeah that number is definitely wrong.

The best fit I can find is that 128 bits is roughly enough to address every cubic meter in a sphere slightly larger than Neptune's orbit.

64 bits is way too big for a single dimension, but way too small for three dimensions.

128 bits is enough to allocate 170 unique addresses to every milligram of matter in the Solar System (~99.9% of which is the Sun, about 2 * 10^33 grams). I think that's a more useful measure than the volume of the Solar System, which after all is mostly empty space. After all, what's the smallest (lightest) device which can benefit from having its own IPv6 address?

https://en.wikipedia.org/wiki/Solar_System suggests 40AU is a reasonable radius (call it 6 billion km). Which gives a spherical volume of 9.05x10^38 m^3 giving you one address per 4.16x10^10 km^3.

You could consider it a disc that just encapsulates the sun - then the volume is 1.57x10^35 m^3 and your address gets you a mere 8.5x10^6 km^3.