Consider a small surface area of 1 mm^2 at the top of a mercury...

The pressure inside the mercury drop is given by $P = P_0 + \dfrac{2T}{R}$, where $P_0$ is the atmospheric pressure, $T$ is the surface tension, and $R$ is the radius of the drop.

Given:
$P_0 = 1.0 \times 10^5 \text{ Pa}$
$T = 0.465 \text{ N/m}$
$R = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}$

The excess pressure inside the drop is:
$\Delta P = \dfrac{2T}{R} = \dfrac{2 \times 0.465}{4.0 \times 10^{-3}} = 232.5 \text{ Pa}$

The total pressure of the mercury just inside the surface is:
$P = P_0 + \Delta P = 100000 + 232.5 = 100232.5 \text{ Pa}$

The force exerted by the mercury below the small surface area $A = 1 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2$ is due to this internal pressure:
$F = P \times A = 100232.5 \times 1.0 \times 10^{-6} = 0.1002325 \text{ N}$

Rounding to five decimal places, we get $F \approx 0.10023 \text{ N}$.