a
Rated power $\&$ voltage gives resistance
$\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50}$
$\mathrm{R}=800$
$\mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800}$
$\mathrm{P}=12.5 \text { watt }$
Hence option $1$ is correct.