To solve this problem, we need to find the pressure inside the soap bubble and equate it to the pressure at a depth $Z_0$ below the water surface.$\\$The excess pressure inside a soap bubble is given by:$\\ P_{excess} = \frac{4T}{r} \\$where $T$ is the surface tension and $r$ is the radius of the bubble.$\\$Given:$\\ T = 2.5 \times 10^{-2} \,$N/m$\\ r = 1 \,$mm$= 1 \times 10^{-3} \,$m$\\$Substitute these values into the formula for excess pressure:$\\ P_{excess} = \frac{4 \times 2.5 \times 10^{-2}}{1 \times 10^{-3}} \\ P_{excess} = \frac{10 \times 10^{-2}}{10^{-3}} \\ P_{excess} = 10^3 \,$N/m$^2 \\$The pressure at a depth $Z_0$ in water is given by:$\\ P_{depth} = \rho g Z_0 \\$where $\rho = 10^3 \,$kg/m$^3$ is the density of water and $g = 10 \,$m/s$^2. \\$Equate the excess pressure to the pressure at depth:$\\ 10^3 = 10^3 \times 10 \times Z_0 \\ Z_0 = \frac{10^3}{10^4} \\ Z_0 = 0.1 \,$m$\\ Z_0 = 10 \,$cm$\\$Therefore, the value of $Z_0$ is $10 \,$cm$,$ which corresponds to Option 2.