An air bubble of diameter 6mm rises steadily through a solution of...

The bubble is moving with a constant velocity so the net force acting on the body is zero.

Stokes' law defines the drag force which is given by $F = 6\piηrv$

The buoyant force is given by $B = Vρg = \dfrac{4}{3}\pi{r}^{3}ρg$

Hence, it can be written that

$\dfrac{4}{3}\pi{r}^{3}ρg = 6\piηrv ⇒ η = \dfrac{2{r}^{2}ρg}{9v} = \dfrac{2 \times {\left(3 \times {10}^{- 3}\right)}^{2} \times 1750 \times 10}{9 \times 0 . 35 \times {10}^{- 2}} = 10Pas$