The Martian atmospheric pressure is around 600 Pascals. At that pressure the boiling point of water is around 0°C. The average temperature on the surface of Mars is -55°C. The melting point of water is also lower, but this usually changes by a much smaller fraction so we'll assume you want to get it to just below boiling. So you only need to increase the temperature by 55°C to be able to store it in liquid form (and then slowly raise its temperature and pressure thereafter). The specific heat capacity of water is around 2,000J/Kg.K. That means you need 2,000 Joules of energy to raise 1kg of water by 1°C. So to raise 1kg by 55°C you need: 2,000x55 = 110,000 Joules. On top of that you need to provide the energy for the enthalpy of fusion. for water this is 334,000J/kg, bringing your total value up to 444,000J/kg. Now you just need to know how much water you need per day.
Aside from that though there's one more useful calculation; the size of solar array required for melting 1kg. The solar flux at Mars is around 557W/m2. Assuming you get that for around half the day (if you have tracking panels) then you're looking at 557 * 12 * 3600 (Martian day is pretty much the same as Earth) = 24,062,400 Joules/m2/day. Solar panels efficiencies are pretty bad, around 14% is typical: 24,062,400 *0.14 = 3,368,736. That means you need 444,000/3,368,736 = 0.1318 m2/kg/day of solar array to melt ice.
This is achievable by robots, but giving an example is a little to broad to answer here.
I hope my thesis has this density of references.