Two solid spheres, A and B, are made of the same material and are...

According to Stokes' law, the viscous drag force on a spherical body is:
$\mathrm{F} = 6 \pi \eta r v$

The terminal velocity $v_t$ of a sphere falling through a viscous fluid is given by:
$v_t = \dfrac{2}{9} \dfrac{r^2 (\rho - \sigma) g}{\eta}$

This shows that the terminal velocity is directly proportional to the square of the radius ($v_t \propto r^2$).

Substituting $v_t$ into the drag force equation gives:
$\mathrm{F} \propto r \times r^2$
$\mathrm{F} \propto r^3$

Alternatively, at terminal velocity, the net force is zero, meaning the viscous drag force balances the effective weight (weight minus buoyant force). Since effective weight is proportional to the volume of the sphere, $\mathrm{F} \propto r^3$.

Given the ratio of radii $r_A : r_B = 1:2$, the ratio of the viscous drag forces is:
$\dfrac{\mathrm{F}_A}{\mathrm{F}_B} = \left(\dfrac{r_A}{r_B}\right)^3 = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$

Answer: $1:8$