Answer
$q=16.44\ kJ/kg$
Work Step by Step
The energy balance for the system is:
$q-w_b-w_{sh}=0$
The boundary work is given by:
$w_b=RT_1\ln(\frac{v_2}{v_1})$
With $R=0.2870\ kJ/kg.K, T_1=17°C=290K, v_2/v_1=3$
$w_b=91.44\ kJ/kg$
And since $w_{sh}=-75\ kJ/kg$:
$q=16.44\ kJ/kg$
