An air bubble of radius R is rising in a liquid of density and...

The terminal velocity $v$ of a spherical object of radius $r$ and density $\sigma$ moving in a liquid of density $\rho$ and viscosity $\eta$ is given by:

$v = \dfrac{2r^2g}{9\eta}|\sigma - \rho|$

For the rising air bubble, the density of air is negligible ($\sigma \approx 0$). The effective density difference driving the motion is $|0 - \rho| = \rho$.
Thus, its terminal speed is:
$v_1 = \dfrac{2R^2g\rho}{9\eta}$

For the falling solid ball, its density is $\sigma = 3\rho$. The effective density difference is $|3\rho - \rho| = 2\rho$.
Thus, its terminal speed is:
$v_2 = \dfrac{2R^2g(2\rho)}{9\eta}$

The ratio of their terminal speeds is:
$\dfrac{v_1}{v_2} = \dfrac{\rho}{2\rho} = \dfrac{1}{2}$

Therefore, the ratio is $1 : 2$.

Answer: $1 : 2$