Answer
$C_{cup}=1.395J/gC^{\circ}$
Work Step by Step
As the heat released by the coffee is given as
$q_{coffee}=m_{coffee}.C_{coffee}.\Delta T$
We plug in the known values to obtain:
$q_{coffee}=(250)(4.184)(100-800)$
$q_{coffee}=20920J$
and the heat gained by the cup material is given as
$q_{cup}=m_{cup}.C_{cup}\Delta T_{cup}$
$\implies q_{cup}=(250)C_{cup}(80-20)$
$\implies q_{cup}=15000 .C_{cup}gC^{\circ}$
As the heat gained by the cup material is equal to the heat lost by the coffee, therefore
$20920J=(15000).C_{cup}gC^{\circ}$
This simplifies to:
$C_{cup}=1.395J/gC^{\circ}$
