d
The two heaters with different power are connected in series,
$P 1=\frac{V^{2}}{R 1}$ and $P 2=\frac{V^{2}}{R 2}$
therefore,
$R 1=\frac{V^{2}}{P 1}$ and $R 2=\frac{V^{2}}{P 2}$
in series combination
$R_{e q}=R 1+R 2$
Hence,
$P_{e q}=\frac{V^{2}}{P 1}+\frac{V^{2}}{P 2}$
$P_{e q}=\left(\frac{1}{2}\right) V^{2}$
$P_{e q}=\frac{1}{2}(400)=200 W$