The CFSE for octahedral ${\left[{\mathrm{CoCl}}_6\right]}^{4-}$ complex is $18000 {\mathrm{cm}}^{-1}$. Then $∆$ for tetrahedral ${\left[{\mathrm{CoCl}}_4\right]}^{2-}$ will be:

Given, ${\mathrm{CFSE}}_{{\mathrm{\Delta }}_0}=18000{\mathrm{cm}}^{-1}$ 

The relationship between the crystal field stabilization energies for octahedral and tetrahedral field is 

${\mathrm{\Delta }}_t=\frac{4}{9}{\mathrm{\Delta }}_0$

Substitute value in the preceding expression.

${\left[{\mathrm{CoCl}}_4\right]}^{2-}=\frac{4}{9}\times 18000{\mathrm{cm}}^{-1}=8000{\mathrm{cm}}^{-1}$.