Answer
$W = -1.95 \times 10^3 J$
Work Step by Step
$W_{ba} = 0 $ because $\Delta V = 0$
$W_{ac} = p\Delta V = nR\Delta T$
$W_{ac} = (3)(8.314 J/mol.K)(492 K - 300 K) $
$W_{ac} = 4788.68 J$
$W_{cb} = Q - \Delta U $ and in adiabatic process, $\Delta Q = 0 $ so
$W_{cb} = - \Delta U$
$W_{cb} = - nC_V \Delta T$ and $C_V = \frac{5}{2}R$
$W_{cb} = - n\frac{5}{2}R \Delta T$
$W_{cb} = -(3)(\frac{5}{2} \times 8.314 J/mol.K) (600K - 492 K) $
$W_{cb} = -6734 J$
Total work done is
$W = W_{ba} + W_{ac} + W_{cb} $
$W = 0 + 4788.68 J + (-6734 J)$
$W = -1.95 \times 10^3 J$