You'd need #"286 kJ"# to get from ice at #"-12"^@"C"# to steam at #"121"^@"C"#.
Here are the constants that you must use
Heat of fusion of water: #DeltaH_f# = #334# #"J/g"#
Heat of vaporization of water: #DeltaH_v# = #2257# #"J/g"#
Specific heat of ice: #c# = #2.09# #"J"#/#"g"^@"C"#;
Specific heat of water: #c# = #4.18# #"J"#/#"g"^@"C"#;
Specific heat of steam: #c# = #2.09# #"J"#/#"g"^@"C"#;
Notice that you must go from ice to vapor, which means that you must go through two phase changes - solid to liquid and liquid to vapor.
The steps that describe your specific transition are
1. Going from ice at #"-12"^@"C"# to ice at #"0"^@"C"#
#q_1 = m * c_("ice") * DeltaT_1 = "93.0 g" * 2.09"J"/("g" * ^@"C") * (0 - (-12))^@"C"#
#q_1 = "2332.44 J"#
2. Going from ice at #"0"^@"C"# to water at #"0"^@"C"#
#q_2 = m * DeltaH_f = "93.0 g" * 334"J"/"g" = "31062 J"#
3. Going from water at #"0"^@"C"# to water at #"100"^@"C"#
#q_3 = m * c_("water") * DeltaT_3#
#q_3 = "93.0 g" * 4.18"J"/("g" * ^@"C") * (100 - 0)^@"C" = "38874 J"#
4. Going from water at #"100"^@"C"# to steam at #"100"^@"C"#
#q_4 = m * DeltaH_v = "93.0 g" * 2257"J"/"g" = "209901 J"#
5. Going from steam at #"100"^@"C"# to steam at #"121"^@"C"#
#q_5 = m * c_("steam") * DeltaT_5#
#q_5 = "93.0 g" * 2.09"J"/("g" * ^@"C") * (121 - 100)^@"C" = "4081.77 J"#
The total amount of energy required to go from ice at #"-12"^@"C"# to steam at #"121"^@"C"# will be the sum of all the calculated energies
#q_("total") = q_1 + q_2 + q_3 + q_4 + q_5#
#q_("total") = "286251.21 J"#
In kJ and rounded to three sig figs, the answer will be
#q_("total") = "286.25121 kJ" = "286 kJ"#
