As given that same tension is applied on the wires, for extension we use young's law equation i.e. $\gamma = \frac{\text{stress}}{\text{strain}}$ , where $\gamma$ is the young's modulus.
Now as we know that stress $= \frac{F}{A}$ , where F is the force and A is the area. Then we know $A = \pi r^{2}$ where r is the radius, also $r = \frac{D}{2}$ , where D is the diameter.
$So A = \frac{\pi D^{2}}{4}. Now we know the formula of strain i.e. strain = \frac{\Delta L}{L}, where \Delta L is the elongation and L is the actual length. Now substituting the equations of stress and strain in young's modulus and solving the equation for \Delta L we get: \gamma = \frac{4F}{\pi D^{2}} \frac{L}{\Delta L} Now we find \Delta L i.e. \Delta L = \frac{4F}{\pi D^{2}} \frac{L}{\gamma}. As we can see that \Delta L \propto \frac{L}{D^{2}} i.e. elongation is directly proportional to \frac{L}{D^{2}}. So, the wire which has the largest extension has the largest \frac{L}{D^{2}} value. Now we check each and every option one by one and find which has the largest \frac{L}{D^{2}} value For wire A, \frac{L}{D^{2}} = \frac{50}{0.5^{2}} = 20000 For wire B, \frac{L}{D^{2}} = \frac{100}{0.1^{2}} = 10000 For wire C, \frac{L}{D^{2}} = \frac{200}{0.2^{2}} = 5000 For wire D, \frac{L}{D^{2}} = \frac{300}{0.3^{2}} = 3333.33.$
Thus, wire A has the largest extension.