To solve this problem, we need to find the potential of the bigger drop formed by the combination of 27 smaller drops.$\\$Given:$\\$• Each small drop is charged at $220$ V.$\\$• Total number of small drops $n = 27. \\$Let's denote:$\\$• $V_s = 220$ V as the potential of each small drop.$\\$• $V_b$ as the potential of the bigger drop.$\\$• $r_s$ as the radius of each small drop.$\\$• $r_b$ as the radius of the bigger drop.$\\$The volume of a single small drop is $\frac{4}{3} \pi r_s^3. \\$The total volume of 27 small drops is $27 \times \frac{4}{3} \pi r_s^3. \\$When these drops combine to form a bigger drop, the volume of the bigger drop is:$\\ \frac{4}{3} \pi r_b^3 = 27 \times \frac{4}{3} \pi r_s^3. \\$Simplifying, we find:$\\ r_b^3 = 27 r_s^3. \\$Taking the cube root of both sides:$\\ r_b = 3 r_s. \\$The potential of a charged sphere is given by $V = \frac{kQ}{r},$ where $Q$ is the charge and $r$ is the radius.$\\$Since the potential of each small drop is $V_s = \frac{kQ_s}{r_s} = 220$ V,$\\$the charge on each small drop is $Q_s = \frac{220 r_s}{k}. \\$The total charge on the bigger drop is $Q_b = 27 Q_s. \\$Substituting for $Q_s: \\ Q_b = 27 \times \frac{220 r_s}{k}. \\$The potential of the bigger drop is $V_b = \frac{kQ_b}{r_b}. \\$Substituting for $Q_b$ and $r_b: \\ V_b = \frac{k \times 27 \times \frac{220 r_s}{k}}{3 r_s}. \\$Simplifying:$\\ V_b = \frac{27 \times 220 r_s}{3 r_s}. \\ V_b = 9 \times 220. \\ V_b = 1980$ V.$\\$Therefore, the potential of the bigger drop is $1980$ V, which corresponds to Option 4.