To find the maximum possible percentage error in the measurement of density, we first need to understand the formula for density.$\\$The density $\rho$ of a wire is given by:$\\ \rho = \frac{m}{V} \\$where $m$ is the mass and $V$ is the volume of the wire.$\\$For a cylindrical wire, the volume $V$ is given by:$\\ V = \pi r^2 l \\$where $r$ is the radius and $l$ is the length of the wire.$\\$Thus, the density becomes:$\\ \rho = \frac{m}{\pi r^2 l} \\$To find the percentage error in density, we use the formula for the propagation of errors:$\\ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \\$Given values:$\\ m = 0.4$ g, $\Delta m = 0.002$ g$\\ r = 0.3$ mm, $\Delta r = 0.001$ mm$\\ l = 5$ cm, $\Delta l = 0.02$ cm$\\$Convert all measurements to consistent units (e.g., cm):$\\ r = 0.03$ cm, $\Delta r = 0.001$ cm$\\$Calculate the relative errors:$\\ \frac{\Delta m}{m} = \frac{0.002}{0.4} = 0.005 \\ \frac{\Delta r}{r} = \frac{0.001}{0.03} \approx 0.0333 \\ \frac{\Delta l}{l} = \frac{0.02}{5} = 0.004 \\$Substitute these into the error propagation formula:$\\ \frac{\Delta \rho}{\rho} = 0.005 + 2 \times 0.0333 + 0.004 \\$Simplify:$\\ \frac{\Delta \rho}{\rho} = 0.005 + 0.0666 + 0.004 \\ \frac{\Delta \rho}{\rho} = 0.0756 \\$Convert to percentage error:$\\$Percentage error$= 0.0756 \times 100\% = 7.56\% \\$The maximum possible percentage error in the measurement of density is approximately $7.6\%. \\$This corresponds to Option 3.