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

For a small area $\Delta A$ on the surface of a liquid drop in equilibrium, the outward force due to the pressure inside the drop is balanced by the inward force due to atmospheric pressure and the inward resultant force exerted by the surrounding surface due to surface tension.

Let $P_{in}$ be the pressure inside the drop and $P_0$ be the atmospheric pressure. The excess pressure is given by:

$P_{ex} = P_{in} - P_0 = \dfrac{2T}{R}$

The force exerted by the surrounding mercury surface on the small area $\Delta A$ is the resultant of the surface tension forces, which is directed radially inward. For equilibrium of the area $\Delta A$:

$P_{in} \Delta A = P_0 \Delta A + F_s$

$F_s = (P_{in} - P_0) \Delta A = P_{ex} \Delta A = \dfrac{2T}{R} \Delta A$

Given:
$T = 0.465 \text{ N/m}$
$R = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}$
$\Delta A = 1 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2$

Substituting the values:

$F_s = \dfrac{2 \times 0.465}{4.0 \times 10^{-3}} \times 1.0 \times 10^{-6}$

$F_s = \dfrac{0.93}{4.0 \times 10^{-3}} \times 10^{-6} = 232.5 \times 10^{-6} \text{ N}$

$F_s = 0.0002325 \text{ N} \approx 0.00023 \text{ N}$