Falling vertically in glycerine, a metal sphere has a mass of 50 mg...

The hydrostatic force exerted by the fluid on the sphere is the buoyant force acting on it.

The formula for buoyant force is given by:
$F_B = V \rho g$

Where:
$V$ is the volume of the sphere $= \dfrac{4}{3} \pi r^3$
$\rho$ is the density of the fluid (glycerine) $= 1260 \text{ kg/m}^3$
$g$ is the acceleration due to gravity $= 9.8 \text{ m/s}^2$
$r$ is the radius of the sphere $= 1 \text{ mm} = 10^{-3} \text{ m}$

Substituting the values into the formula:
$F_B = \dfrac{4}{3} \pi (10^{-3})^3 \times 1260 \times 9.8$

$F_B = \dfrac{4}{3} \times \dfrac{22}{7} \times 10^{-9} \times 1260 \times 9.8$

$F_B = 4 \times 22 \times 60 \times 9.8 \times 10^{-9}$

$F_B = 5280 \times 9.8 \times 10^{-9}$

$F_B = 51744 \times 10^{-9} \text{ N}$

$F_B = 5.1744 \times 10^{-5} \text{ N}$

Rounding off to two significant figures, we get:
$F_B \approx 5.2 \times 10^{-5} \text{ N}$