A uniform rod of length L has a mass per unit length $\mathrm{\lambda }$ and area of cross-section A.
The elongation in the rod is l due to its own weight.  If it is suspended from the ceiling of a room, the Young's modulus of the rod is

Young's modulus, E, is defined as the ratio of the applied tensile stress to the corresponding strain in the material. In this case, the tensile stress is caused by the weight of the rod itself and can be calculated as:

$\mathrm{Stress} =\frac{\mathrm{Force}}{\mathrm{Area}} =\frac{\mathrm{\lambda gL}}{\mathrm{A}}$

where g is the acceleration due to gravity. The strain, ε, is given by the change in length of the rod divided by its original length:

$\mathrm{Strain}=\frac{∆\mathrm{L}}{\mathrm{L}}=\frac{I}{\mathrm{L}}$

Therefore, Young's modulus can be expressed as:

$\mathrm{E} = \frac{\mathrm{Stress}}{\mathrm{Strain}} =\frac{ \mathrm{\lambda g}{\displaystyle \frac{\mathrm{L} }{2}}}{{\displaystyle \frac{\mathrm{A}l}{\mathrm{L}} }} =\frac{ {\mathrm{\lambda gL}}^2 }{2\mathrm{A}l}$

Hence the correct answer is $\frac{{\mathrm{\lambda gL}}^2}{2\mathrm{A}l}.$