1. For steady rise, net upward force = viscous drag

Since the air bubble has negligible weight, the effective upward force on it is approximately just the buoyant force:

$F_b = \text{weight of displaced liquid} = \frac{4}{3}\pi r^3 \sigma g$

2. Viscous drag on a spherical bubble

For a small spherical body moving slowly through a viscous liquid, Stokes' law gives:

$F_v = 6\pi \eta r v$

At terminal (steady) speed,

$F_b = F_v$

So,

$\frac{4}{3}\pi r^3 \sigma g = 6\pi \eta r v$

3. Solve for $\eta$

Cancel $\pi$ and one factor of $r$:

$\frac{4}{3} r^2 \sigma g = 6 \eta v$

Hence,

$\eta = \frac{\frac{4}{3} r^2 \sigma g}{6v} = \frac{4 r^2 \sigma g}{18 v} = \frac{2 r^2 \sigma g}{9 v}$

4. Match with options

$\eta = \frac{2 r^2 \sigma g}{9 v}$

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the answer agrees with the stored correct answer.