To solve this problem, we need to relate the current sensitivity and voltage sensitivity of the galvanometer to its resistance.$\\$Given:$\\$• Current sensitivity $= 5$ div/mA$\\$• Voltage sensitivity $= 20$ div/V$\\$Let's denote:$\\$• $S_i = 5$ div/mA as the current sensitivity.$\\$• $S_v = 20$ div/V as the voltage sensitivity.$\\$• $R$ as the resistance of the galvanometer.$\\$The relationship between current sensitivity and voltage sensitivity is given by:$\\ S_v = \frac{S_i}{R} \\$Substitute the known values:$\\ 20 = \frac{5}{R} \\$Rearrange to solve for $R: \\ R = \frac{5}{20} \\ R = \frac{1}{4} \\ R = 0.25$ k$\Omega \\ R = 250 \Omega \\$Therefore, the resistance of the galvanometer is $250 \Omega. \\$This corresponds to Option 1.