A spherical body of radius r and density σ falls freely through a...

The terminal velocity $v_0$ of a spherical body falling through a viscous liquid is given by Stokes' Law:
$v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$
Rearranging the formula to express viscosity $\eta$ in terms of other variables:
$\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$
Since we are told to ignore errors associated with $\sigma$, $\rho$, and $g$, these quantities along with the numerical factor $\frac{2}{9}$ are treated as constants.
The expression for $\eta$ can be written as $\eta = k \cdot \frac{r^2}{v_0}$, where $k = \frac{2(\sigma - \rho)g}{9}$ is a constant.
Taking the natural logarithm on both sides:
$\ln \eta = \ln k + 2\ln r - \ln v_0$
Differentiating both sides to find the relative error:
$\frac{d\eta}{\eta} = 0 + 2\frac{dr}{r} - \frac{dv_0}{v_0}$
For the estimated maximum fractional error, we take the absolute sum of the individual relative errors:
$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$
This matches option (1).