The excess force ($F$) required to lift a flat circular disc from the surface of a liquid is equal to the force due to surface tension acting along the boundary (circumference) of the disc in contact with the liquid.

Formula: $F = T \times L$ Where: $T$ is the surface tension. $L$ is the length of the boundary in contact with the liquid. For a circular disc, $L = \text{Circumference} = 2\pi r$.

Given: Radius, $r = 4.5 \text{ cm} = 4.5 \times 10^{-2} \text{ m}$ Surface Tension, $T = 0.07 \text{ N m}^{-1} = 7 \times 10^{-2} \text{ N m}^{-1}$

Calculation: $F = (7 \times 10^{-2}) \times 2 \times \frac{22}{7} \times (4.5 \times 10^{-2})$ $F = 10^{-2} \times 2 \times 22 \times 4.5 \times 10^{-2}$ $F = 44 \times 4.5 \times 10^{-4}$ $F = 198 \times 10^{-4} \text{ N}$ $F = 1.98 \times 10^{-2} \text{ N}$ To convert to millinewtons (mN): $F = 19.8 \times 10^{-3} \text{ N} = 19.8 \text{ mN}$