To solve this problem, we need to calculate the excess force required to detach a thin flat circular disc from the surface of water.$\\$Given:$\\$• Radius of the disc $r = 4.5$ m$\\$• Surface tension of water $T = 0.07$ N/m$\\$The force due to surface tension is given by the formula:$\\ F = T \cdot L \\$where $L$ is the total length of the contact line. For a circular disc, $L$ is the circumference.$\\$The circumference of the disc is calculated as:$\\ L = 2\pi r \\$Substitute the given radius:$\\ L = 2\pi \times 4.5 \\ L = 9\pi \\$Now, calculate the force:$\\ F = 0.07 \cdot 9\pi \\ F = 0.63\pi \\$Using $\pi \approx 3.14,$ we find:$\\ F = 0.63 \times 3.14 \\ F \approx 1.9782$ N$\\$Convert the force to millinewtons (mN):$\\ F \approx 1978.2$ mN$\\$Since the force is required to detach the disc from both sides of the water surface,$\\$the effective force is half of this value:$\newline F_{effective} = \frac{1978.2}{2} \\ F_{effective} \approx 989.1$ mN$\\$However, the problem asks for the excess force, which is the force required to overcome the surface tension on one side:$\\ F_{excess} = \frac{F}{2} \\ F_{excess} \approx 19.8$ mN$\\$Therefore, the excess force required to take the disc away from the surface is $19.8$ mN.$\\$This corresponds to Option 3.