A steel ball of radius 1.2 mm and density 7800 kg/m³ falls through castor oil of density 960 kg/m³…

Step 1. The terminal speed. $v_t = \dfrac{0.500\ \text{m}}{22.8\ \text{s}} = 0.0219\ \text{m/s}$ $= 2.19$ cm/s.

Step 2. The viscosity. The numerator is $2 \times (1.2 \times 10^{-3}\ \text{m})^2$ $\times 6840\ \text{kg/m}^3$ $\times 9.8\ \text{m/s}^2$ $= 0.193\ \text{kg/s}^2$. The denominator is $9v_t = 9 \times 0.02193\ \text{m/s}$ $= 0.197\ \text{m/s}$. So $\eta = \dfrac{0.1931\ \text{kg/s}^2}{0.1974\ \text{m/s}}$ $= 0.978\ \text{Pa s}$.

Step 3. The Reynolds number. The numerator of $Re = \dfrac{\sigma v_t (2r)}{\eta}$ is $960\ \text{kg/m}^3$ $\times 0.0219\ \text{m/s}$ $\times 2.4 \times 10^{-3}\ \text{m}$ $= 0.0505\ \text{Pa s}$. So $Re = \dfrac{0.0505\ \text{Pa s}}{0.978\ \text{Pa s}}$ $= 0.052$.

Is Stokes' law fair? Yes. The Reynolds number is far below $1$, the ball is small, and the jar is wide. The flow around the ball is smooth.