The terminal velocity of a small spherical object moving under the action of gravity through a fluid medium is given by Stokes' Law, which is stated as:

$v = \frac{2}{9} \frac{r^2 g (\rho_p - \rho_f)}{\eta}$,

where:

• $v$ is the velocity of the object (in this case, the air bubble),
• $r$ is the radius of the object,
• $g$ is the acceleration due to gravity,
• $\rho_p$ is the density of the object (negligible in this case, as it's an air bubble),
• $\rho_f$ is the density of the fluid (the solution), and
• $\eta$ is the coefficient of viscosity of the fluid.

Since we are neglecting the density of the air bubble, the formula simplifies to:

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

Rearranging for $\eta$, we get:

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

Given that $r = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$, $g = 10 \, \text{ms}^{-2}$, $\rho_f = 1750 \, \text{kg/m}^{3}$, and $v = 0.35 \, \text{cm/s} = 0.35 \times 10^{-2} \, \text{m/s}$, we can substitute these values into the formula to find $\eta$:

$\eta = \frac{2}{9} \frac{(3 \times 10^{-3})^2 \times 10 \times 1750}{0.35 \times 10^{-2}} = 10 \, \text{Pas}$.

Therefore, the coefficient of viscosity of the solution is $10 \, \text{Pas}$.