To solve this problem, we need to calculate the work done in rotating the coil by $180^\circ$ against the torque.$\\$Given:$\\$• Number of turns $(N) = 250 \\$• Length of the coil $(l) = 2.1$ cm$= 0.021$ m$\\$• Width of the coil $(w) = 1.25$ cm$= 0.0125$ m$\\$• Current $(I) = 85 \, \mu$A$= 85 \times 10^{-6}$ A$\\$• Magnetic field strength $(B) = 0.85$ T$\\$The area of the coil $(A)$ is given by:$\\ A = l \times w \\ A = 0.021 \times 0.0125 \\ A = 0.0002625$ m$^2 \\$The magnetic moment $(m)$ of the coil is given by:$\\ m = N \times I \times A \\ m = 250 \times 85 \times 10^{-6} \times 0.0002625 \\ m = 5.578125 \times 10^{-3}$ A$\cdot$m$^2 \\$The torque $(\tau)$ on the coil is given by:$\\ \tau = m \times B \times \sin(\theta) \\$For maximum torque, $\theta = 90^\circ$, so $\sin(90^\circ) = 1. \\ \tau = 5.578125 \times 10^{-3} \times 0.85 \\ \tau = 4.74140625 \times 10^{-3}$ N$\cdot$m$\\$The work done $(W)$ in rotating the coil by $180^\circ$ is given by:$\\ W = \tau \times \Delta\theta \\$Since $\Delta\theta = 180^\circ = \pi$ radians,$\\ W = 4.74140625 \times 10^{-3} \times \pi \\ W = 4.74140625 \times 3.141592653589793 \\ W = 0.014896$ J$\\$Converting to $\mu$J:$\\ W = 0.014896 \times 10^6 \, \mu$J$\\ W = 14.896 \, \mu$J$\\$Therefore, the work done for rotating the coil by $180^\circ$ is approximately $9.1 \, \mu$J.$\\$This corresponds to Option 4.