The diameter of an air bubble which was initially 2mm , rises...

As the bubble is rising steadily the net force acting on it will be zero.

(Because of density of air is negligible, so the value of $mg$ can be neglected)

So, the force of buoyancy $B$ is equal to the weight of the displaced liquid acting upward is equal to the viscous force acting downward, i.e., $B = F$.

$⇒ \dfrac{4\pi}{3}{R}^{3}ρg = 6\piηRv$

Putting $R = 1mm = {10}^{- 3}m$, $ρ = 1 . 75 \times {10}^{3}kg{m}^{- 3}$, g = 10m{s}^{- 2} and $v = 0 . 35 \times {10}^{- 2}m{s}^{- 1}$

$η = \dfrac{10}{9} = 1 \cdot 11$

In SI unit, $η = 11$ poise \left(CGS\right).