a
$\left(T_{2}\right) T_{\sin k}=200 \; K$
$\left(T_{1}\right) T_{\text {Reservoir }}=527+273=800 \; K$
$W=12000 K J=12 \times 10^{6} \; J$
$Q_{1}=$ ?
$\eta=1-\frac{T_{2}}{T_{1}}=\frac{W}{Q_{1}}=1-\frac{200}{800}=\frac{12 \times 10^{6}}{Q_{1}}$
$\frac{3}{4}=\frac{12 \times 10^{6}}{Q_{1}}=Q_{1}=16 \times 10^{6} J$