The correct order of atomic radii is:

First, let us recall the general rule for atomic size: as we move from left to right in a period of the periodic table, effective nuclear charge increases, so atomic radius decreases; as we move down a group, addition of extra shells makes the radius increase. Mathematically we write this qualitative relationship as $$r \propto \dfrac{1}{Z_{\text{eff}}}$$ where $$r$$ is the atomic radius and $$Z_{\text{eff}}$$ is the effective nuclear charge experienced by the outer-most electrons.

Now we examine the four elements appearing in the options.

$$\text{Ce (Z = 58)},\; \text{Eu (Z = 63)},\; \text{Ho (Z = 67)}$$ all belong to the lanthanide series (4f-block). Inside that series, as the atomic number increases, the outer electrons are added to the 4f subshell. The 4f electrons shield nuclear charge very poorly. Hence, with increasing $$Z$$ the effective nuclear charge $$Z_{\text{eff}}$$ rises appreciably and the radius should normally shrink; this slow but steady shrinkage is called the lanthanide contraction.

However, the experimental data show two noticeable bulges in the contraction curve: one at europium (Eu) and another at ytterbium (Yb). The reason is the unusual stability of a half-filled 4f7 configuration (Eu) and a completely filled 4f14 configuration (Yb). The stability reduces the tendency of the 4f electrons to penetrate closer to the nucleus, so the shrinkage momentarily slows down. Stated algebraically, for Eu we find a larger metallic radius:

$$r_{\text{Eu}} \approx 199\ \text{pm}, \qquad r_{\text{Ce}} \approx 182\ \text{pm}, \qquad r_{\text{Ho}} \approx 178\ \text{pm}.$$

Thus within the lanthanides quoted we have the relation

$$r_{\text{Eu}} > r_{\text{Ce}} > r_{\text{Ho}}.$$

Next, compare any of these radii to that of nitrogen, $$\text{N (Z = 7)}$$. Nitrogen lies in the second period, p-block, with only two principal shells (K and L). Because there are so few shells, the distance of the valence electrons from the nucleus is very small. Its covalent (atomic) radius is about $$r_{\text{N}} \approx 70\ \text{pm}$$, far smaller than any of the metallic radii listed for the lanthanides.

Hence, when nitrogen is included, the complete decreasing-size order becomes

$$r_{\text{Eu}} > r_{\text{Ce}} > r_{\text{Ho}} > r_{\text{N}}.$$

This sequence matches exactly the arrangement given in Option D.

Hence, the correct answer is Option D.