The change in entropy of the surroundings is very easy, since the process is essentially isothermal (the temperature of the surroundings does not change during the process). The reversible path for the surroundings is just an isothermal reversible path in which the heat flowing out of the surroundings and into the ice does so at constant temperature. (1) What is the total heat flow into the surroundings? (2) What is the temperature. Divide (1) by (2) to get ΔS of the surroundings.
When you have an irreversible process, to calculate the entropy of the system and of the surroundings you have to use a different reversible path for each. In the case of the ice going from 271K to 273K, you have to use a quasi-static path in which the ice is in thermal contact with a body at an infinitessimally higher temperature that keeps increasing slowly as the ice absorbs heat flow, Q. For the surroundings you would use a quasi-static path in which the surroundings are in contact with a body at an infinitessimally lower temperture until Q heat flow leaves the surroundings.