For $3\mathrm{rd}$ ionisation energy, the electron must be removed from $+2$ oxidation state of given metals. The electronic configuration of $\mathrm{Ce}$ is

$4{\mathrm{f}}^{1}5{\mathrm{d}}^{1}6{\mathrm{s}}^2$

The electronic configuration of ${\mathrm{Ce}}^{2+}$ is

$4{\mathrm{f}}^{1}5{\mathrm{d}}^1$

The electronic configuration of $\mathrm{Nd}$ is

$4{\mathrm{f}}^{4}6{\mathrm{s}}^2$

The electronic configuration of ${\mathrm{Nd}}^{2+}$ is

$4{\mathrm{f}}^4$

The electronic configuration of $\mathrm{Eu}$ is

$4{\mathrm{f}}^{7}6{\mathrm{s}}^2$

The electronic configuration of ${\mathrm{Eu}}^{2+}$ is

$4{\mathrm{f}}^7$

The electronic configuration of $\mathrm{Dy}$ is

$4{\mathrm{f}}^{10}6{\mathrm{s}}^2$

The electronic configuration of ${\mathrm{Dy}}^{2+}$ is $4{\mathrm{f}}^{10}$ $\mathrm{Eu}$ will have the maximum $3\mathrm{rd}$ ionisation enthalphy, because the electron has to be removed from stable half-filled $\left(4{\mathrm{f}}^7\right){\mathrm{Eu}}^{2+}$ configuration, whereas $\mathrm{Ce}$ will have the lowest $3\mathrm{rd}$ ionisation enthalpy as the electron has to be removed from $5\mathrm{d}$-orbital.