1. Use Stokes' law for terminal velocity

For a small sphere rising steadily in a viscous liquid, the terminal speed is

$v = \frac{2 r^2 g (\rho_{\text{liquid}}-\rho_{\text{bubble}})}{9\eta}$

Since density of air is neglected,

$\rho_{\text{bubble}} \approx 0$

So,

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

Hence,

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

2. Convert given quantities into SI units

• Diameter of bubble $= 6\text{ mm}$

$r = 3\text{ mm} = 3\times 10^{-3}\text{ m}$

• Speed of rise $= 0.35\text{ cm/s}$

$v = 0.35\times 10^{-2} = 3.5\times 10^{-3}\text{ m/s}$

• Density of solution:

$\rho = 1750\text{ kg/m}^3$

• $g = 10\text{ m/s}^2$

3. Substitute into the formula

$\eta = \frac{2(3\times 10^{-3})^2(10)(1750)}{9(3.5\times 10^{-3})}$

First,

$(3\times 10^{-3})^2 = 9\times 10^{-6}$

So,

$\eta = \frac{2\times 9\times 10^{-6}\times 10\times 1750}{9\times 3.5\times 10^{-3}}$

Cancel $9$:

$\eta = \frac{2\times 10^{-6}\times 10\times 1750}{3.5\times 10^{-3}}$

Now,

$2\times 10\times 1750 = 35000$

Thus,

$\eta = \frac{35000\times 10^{-6}}{3.5\times 10^{-3}}$

$\eta = \frac{3.5\times 10^{-2}}{3.5\times 10^{-3}} = 10$

4. Final answer

$\boxed{10\ \text{Pa·s}}$

So the required integer answer is

$\boxed{10}$

5. Comparison with stored correct answer

Stored correct answer = $10$

Our derived answer = $10$

They match.