Answer
b) 6.95 kJ/°C
Work Step by Step
As the reaction gives off heat, $q_{rxn}$ is negative.
Heat capacity of the calorimeter
$C_{cal}= \frac{q_{cal}}{\Delta T}=\frac{-q_{rxn}}{\Delta T}=\frac{-(-26.42\,kJ)}{(27.20-23.40)^{\circ}C}$
$=6.95\,kJ/^{\circ}C$
