Ionization enthalpy is the energy required to remove the electron from the outer most orbit.

Electronic configuration :

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^2} \cr} $$

After first ionization enthalpy the electronic configuration will be,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^1} \cr} $$

Now in 2nd ionization enthalpy one more electron will be removed. And after second ionization enthalpy electronic configuration will be,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^4} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6} \cr} $$

For $$V(23), Mn (25)$$ and $$Fe (26)$$ electron is removed from $$4S$$ orbital but for $$Cr (24)$$ electron is removed from $$3d$$ orbital.

We know distance of $$4S\,\, > \,\,3d$$ from the nucleus. So, the attraction on the electrons of $$45$$ shell is less compared to $$3d$$ shell by the nucleus. So, the removed of electron will be easier for the electrons of $$4S$$ shell than $$3d$$ shell.

So, for $$Cr(24)$$ second ionization is more than other as electron is removed from $$3d$$ shell.

And for $$Cr(24)$$ outer most shell $$''3d''$$ was half filled after $$1st$$ ionization enthalpy, so $$3d$$ shell was stable. So removal of electron will be difficult from stable shell. That is why also $$2nd$$ ionization enthalpy of $$Cr(24)$$ is higher compare to other given atoms.