If an air bubble of diameter 2 mm rises steadily through a liquid...

The terminal velocity of an air bubble rising in a liquid is given by Stokes' law:

$v = \dfrac{2}{9} \dfrac{r^2 \rho g}{\eta}$

Where:
$r = \dfrac{D}{2} = 1 \text{ mm} = 10^{-3} \text{ m}$
$\rho = 2000 \text{ kg/m}^3$
$g = 10 \text{ m/s}^2$
$v = 0.5 \text{ cm/s} = 5 \times 10^{-3} \text{ m/s}$

Rearranging the formula for the coefficient of viscosity $\eta$:

$\eta = \dfrac{2}{9} \dfrac{r^2 \rho g}{v}$

Substituting the given values:

$\eta = \dfrac{2}{9} \times \dfrac{(10^{-3})^2 \times 2000 \times 10}{5 \times 10^{-3}}$

$\eta = \dfrac{2}{9} \times \dfrac{10^{-6} \times 20000}{5 \times 10^{-3}}$

$\eta = \dfrac{2}{9} \times \dfrac{2 \times 10^{-2}}{5 \times 10^{-3}}$

$\eta = \dfrac{2}{9} \times 4 = \dfrac{8}{9} \text{ Pa}\cdot\text{s}$

To convert the viscosity from SI units (Pa$\cdot$s) to CGS units (Poise), we multiply by $10$:

$\eta = \dfrac{8}{9} \times 10 = \dfrac{80}{9} \text{ Poise}$

$\eta \approx 8.88 \text{ Poise}$

The closest given option is $8.8$.

Answer: $8.8$