To solve this problem, we need to determine the elastic potential energy stored in the wire when it is extended.$\\$Given:$\\$• Mass of the block $M \\$• Original length of the wire $L \\$• Extended length of the wire $(L + l) \\$The extension of the wire is $l. \\$The elastic potential energy stored in a stretched wire is given by:$\\ U = \frac{1}{2} \cdot \frac{F^2}{Y} \cdot \frac{L}{A} \\$Where:$\\$• $F$ is the force applied (equal to the weight of the block, $Mg) \\$• $Y$ is the Young's modulus of the wire$\\$• $A$ is the cross-sectional area of the wire$\\$However, we can also express the potential energy in terms of the extension $l$ as follows:$\\ U = \frac{1}{2} \cdot F \cdot l \\$Since the force $F$ is the weight of the block, we have:$\\ F = Mg \\$Substitute $F = Mg$ into the expression for $U: \\ U = \frac{1}{2} \cdot Mg \cdot l \\$Thus, the elastic potential energy stored in the extended wire is:$\\ U = \frac{1}{2} Mgl \\$Therefore, the correct option is Option 3.