A small ball of mass M and density ρ is dropped in a viscous liquid...

The buoyant force is given by

$B = V{ρ}_{0}g$

where $V$ is the volume of the ball. The weight of the ball

$W = Mg = Vρg$, is acting in the opposite direction of the buoyant force . Hence, at terminal velocity

$Mg - B = {F}_{v}$

$⇒ B + {F}_{v} = Mg$

$⇒ {F}_{v} = Vρg - V{ρ}_{0}g ⇒ {F}_{v} = ρVg ( 1 - \dfrac{{ρ}_{0}}{ρ} ) ⇒ {F}_{v} = Mg ( 1 - \dfrac{{ρ}_{0}}{ρ} )$