The final temperature of the olive oil will be #"33.3"^@"C"#.
So, you know that you have a mass of olive oil, 55.0 g to be precise, to which you add 877 calories of energy in the form of heat.
A substance's specific heat tells you how much heat is needed to increase the temperature of 1 gram of that substance by 1 degree Celsius. Mathematically, the relationship between supplied heat and increase in a substance's temperature is
#q = m * c * DeltaT#, where
#q# - the amount of heat supplied;
#m# - the mass of the substance - in your case olive oil;
#c# - the olive oil's specific heat, 2.19 cal/g C;
#DeltaT# - the change in temperature - defined as the difference between the final and the initial temperature of the substance.
So, in order to determine by how much the temperature of the olive oil increased, you must solve for #DeltaT# in the above equation
#DeltaT = q/(m * c) = "877 cal"/("55.0 g" * 2.19"cal"/("g" * ^@"C")#
#DeltaT = 7.28^@"C"#
Since #DeltaT# is defined as final temperature minus initial temperature, you get
#DeltaT = T_("final") - T_("initial") => T_("final") = DeltaT + T_("initial")#
#T_("final") = 7.28^@"C" + 26.0^@"C" = "33.3"^@"C"#
The answer is rounded to three sig figs.