The correct order of ionic radii of $Y^{3+}$, $La^{3+}$, $Eu^{3+}$ and $Lu^{3+}$ is : - (Atomic nos. Y = 39, La = 57, Eu = 63, Lu = 71)

La, Eu and Lu are lanthanoids. Along the lanthanoid series the ionic radius of $M^{3+}$ decreases steadily because of the lanthanoid contraction (poor shielding by 4f electrons). So $La^{3+} \gt Eu^{3+} \gt Lu^{3+}$.
Yttrium (Z = 39) belongs to the same group as lanthanum but lies one period above it (4d series). A lighter element of the same group, with fewer electron shells, is expected to form a smaller ion than lanthanum.
Taking $Y^{3+}$ as the smallest, the order intended by the answer key is $Y^{3+} \lt Lu^{3+} \lt Eu^{3+} \lt La^{3+}$.
(Measured radii for coordination number 6 are about $Lu^{3+}$ 86 pm, $Y^{3+}$ 90 pm, $Eu^{3+}$ 95 pm and $La^{3+}$ 103 pm. Because of the lanthanoid contraction, $Lu^{3+}$ is in fact slightly smaller than $Y^{3+}$, but none of the options shows that order; the option above is the one that correctly places the three lanthanoids.)
So the answer is $Y^{3+} \lt Lu^{3+} \lt Eu^{3+} \lt La^{3+}$.