Back bonding occurs when there is a transfer of electron density from a filled orbital of one atom to an empty orbital of another atom within a molecule. In $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$, aluminum atoms form a dimer through a bridge involving back bonding. Each aluminum has an empty p-orbital that can accept electron density from the filled p-orbitals of the bridging methyl groups. Therefore, the correct answer is $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$.