One atom of mercury has a mass of $333\times {10}^{-24}$ g. Its atomic weight is 

$atomic weight of an element = N_A(Avogadro\text{'}snumber) \times mass of one atom$

Atomic weight of mercury = $6.023\times {10}^{23}\times 333\times {10}^{-24}$ = $200 a.m.u$

Therefore, Option (A) is correct.