A thin metal disc of radius ‘r’ floats on water surface bends the surface downwards along the perimeter making an angle $\mathrm{\theta }$ with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is

As depicted, the disc is being acted upon by three forces:
weight mg of the metal disc upward at an angle with the vertical buoyancy force, and downward force caused by surface tension (T)
Equating all forces are equal,

$\mathrm{mg}=2\mathrm{\pi rTcos\theta }+\mathrm{W}$

Hence the correct answer is $2\mathrm{\pi rTcos\theta }+\mathrm{W}$