For the electronic transition from ,$\mathrm{n}=2\to \mathrm{n}=1$ which of the following will produce shortest wave length?

According to Rydberg equation:

$\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{\mathrm{z}}^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

As from the equation, the atomic number and wavelength are inversely related. More the atomic number, the shortest will be the wavelength.

The atomic number of lithium is 3 and is the highest of all. So, this nucleus will have the shortest wavelength.