To find the maximum percentage error in the measurement of $X = \frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3},$ we need to consider the percentage errors in each of the variables $A, B, C,$ and $D. \\$The formula for the propagation of errors in a function of multiple variables is given by:$\\ \Delta X = \left| \frac{\partial X}{\partial A} \right| \Delta A + \left| \frac{\partial X}{\partial B} \right| \Delta B + \left| \frac{\partial X}{\partial C} \right| \Delta C + \left| \frac{\partial X}{\partial D} \right| \Delta D \\$However, for percentage errors, we can use the following simplified approach:$\\$If $X = A^m B^n C^p D^q,$ then the percentage error in $X$ is given by:$\\ \Delta X\% = |m| \Delta A\% + |n| \Delta B\% + |p| \Delta C\% + |q| \Delta D\% \\$For the given expression $X = \frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3},$ we have:$\\ m = 2, \, n = \frac{1}{2}, \, p = -\frac{1}{3}, \, q = -3 \\$The percentage errors in $A, B, C,$ and $D$ are $1\%, 2\%, 3\%,$ and $4\%$ respectively.$\\$Substitute these values into the formula:$\\ \Delta X\% = |2| \cdot 1\% + \left|\frac{1}{2}\right| \cdot 2\% + \left|-\frac{1}{3}\right| \cdot 3\% + |-3| \cdot 4\% \\$Calculate each term:$\\ = 2 \cdot 1\% + \frac{1}{2} \cdot 2\% + \frac{1}{3} \cdot 3\% + 3 \cdot 4\% \\ = 2\% + 1\% + 1\% + 12\% \\$Add these values to find the total percentage error:$\\ \Delta X\% = 16\% \\$Therefore, the maximum percentage of error in the measurement of $X$ is $16\%. \\$This corresponds to Option 2.