Two steel spheres, A and B, are falling through the same viscous...

According to Stokes' law, the viscous drag force $F_v$ on a sphere of radius $r$ moving with velocity $v$ in a fluid of viscosity $\eta$ is given by:

$F_v = 6\pi\eta r v$

Since both spheres are falling through the same oil, $\eta$ is constant. Therefore, the viscous drag force is directly proportional to the product of radius and velocity:
$F_v \propto r v$

The ratio of the viscous drag forces is:
$\dfrac{F_A}{F_B} = \left(\dfrac{r_A}{r_B}\right) \times \left(\dfrac{v_A}{v_B}\right)$

Substituting the given ratios:
$\dfrac{F_A}{F_B} = \left(\dfrac{2}{3}\right) \times \left(\dfrac{3}{4}\right) = \dfrac{2}{4} = \dfrac{1}{2}$

Thus, the ratio of the viscous drag forces is $1:2$.

Squaring the ratio of the radii (confusing the drag force proportionality with terminal velocity proportionality) would incorrectly yield $1:3$.

Answer: $1:2$