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

The pressure exerted by the air above the mercury drop is the atmospheric pressure, $P_0 = 1.0 \times 10^5 \text{ Pa}$.

The given surface area on the drop is $A = 1 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2$.

The force exerted by the air on this area is calculated as:

$F = P_0 \times A$

$F = (1.0 \times 10^5) \times (1.0 \times 10^{-6}) = 0.1 \text{ N}$

The surface tension and radius of the drop are not required to find the force exerted by the outside air, as they are used to determine the pressure inside the drop.