b
$R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega$
In series, $R_{e q}=484+484+968 \Omega$
$\therefore P_{e q}=\frac{V^2}{968}=\frac{220 \times 220}{968}=50 \;watt$
In parallel, $R_{\text {eq }}=242 \Omega$
$\therefore P_{e q}=\frac{V^2}{242}=\frac{220 \times 220}{242}=200\; watt$.