First we have to get the $200~W$ as $dB$.

I see they refer to the picowatt for this so

$P_0 = 10 \log_{10}\left(\dfrac{200}{10^{-12}}\right) = 143.01 ~dB$

at $100~m$ the loss due to spherical spreading is

$A = 10\log_{10}\left(\dfrac{1}{4\pi 10^2}\right) = -50.99~dB$

Thus at $100~m$ we have

$P =143.01 - 50.99 = 92.02~dB$

So given the choices I'd choose $92~dB$