To solve this problem, we need to calculate the magnetic dipole moment for each shape.$\\$The magnetic dipole moment $M$ for a current-carrying loop is given by:$\\ M = nIA \\$where $n$ is the number of turns, $I$ is the current, and $A$ is the area of the loop.$\\$Let's analyze each case:$\\$(i) Equilateral Triangle:$\\$• The wire is wound into an equilateral triangle with side $a. \\$• The perimeter of the triangle is $3a. \\$• Since the total length of the wire is $12a,$ the number of turns $n$ is:$\\ n = \frac{12a}{3a} = 4 \\$• The area $A$ of an equilateral triangle with side $a$ is:$\\ A = \frac{\sqrt{3}}{4}a^2 \\$• The magnetic dipole moment $M_1$ for the triangle is:$\\ M_1 = nIA = 4I \left(\frac{\sqrt{3}}{4}a^2\right) = \sqrt{3}Ia^2 \\$(ii) Square:$\\$• The wire is wound into a square with side $a. \\$• The perimeter of the square is $4a. \\$• Since the total length of the wire is $12a,$ the number of turns $n$ is:$\\ n = \frac{12a}{4a} = 3 \\$• The area $A$ of a square with side $a$ is:$\\ A = a^2 \\$• The magnetic dipole moment $M_2$ for the square is:$\\ M_2 = nIA = 3I(a^2) = 3Ia^2 \\$Therefore, the magnetic dipole moments for the equilateral triangle and the square are $\sqrt{3}Ia^2$ and $3Ia^2$ respectively.$\\$This corresponds to Option 1.