To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid:

$ v_T = \frac{2}{9} \frac{(\rho_s - \rho_f) r^2 g}{\eta} $

Where:

$ v_T $ is the terminal velocity,

$ \rho_s $ is the density of the steel ball,

$ \rho_f $ is the density of the fluid,

$ r $ is the radius of the ball,

$ g $ is the acceleration due to gravity,

$ \eta $ is the viscosity of the fluid.

Given:

Diameter of the steel ball = 3.6 mm, therefore radius $ r = 1.8 $ mm = $ 1.8 \times 10^{-3} $ m,

Terminal velocity $ v_T = 2.45 \times 10^{-2} $ m/s,

Density of oil $ \rho_f = 925 $ kg/m$^3$,

Density of steel $ \rho_s = 7825 $ kg/m$^3$,

Acceleration due to gravity $ g = 9.8 $ m/s$^2$.

Rearranging the formula to solve for $\eta$:

$ \eta = \frac{2}{9} \cdot \frac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T} $

Substituting the known values into the equation:

$ \eta = \frac{2}{9} \cdot \frac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}} $

Calculating this, we approximate the viscosity $\eta$ to be:

$ \eta \approx 1.99 \, \text{Pa}\cdot\text{s} $

Thus, the viscosity of the oil is approximately 1.99 Pa·s.