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

(Because of density of air the value of mg can be neglected)

So \(B = F \Rightarrow \frac{4 \pi}{3} R ^{3} \rho g =6 \pi \eta Rv\)

Putting \(R =1\,mm =10^{-3}\,m\)

\(\rho=1.75 \times 10^{3}\,kg / m ^{3}\)

\(g =10\,m / s ^{2}\)

\(v =0.35 \times 10^{-2}\,m / s\)

\(\eta=\frac{10}{9}=1.11\) SI unit \(=11\) poise \(( CGS )\)