In comparison to boron, beryllium has:

Let us first recall that the nuclear charge of an atom is nothing but its atomic number $$Z$$, that is, the number of protons present in the nucleus. Beryllium has atomic number $$Z_{\,\text{Be}} = 4$$ while boron has $$Z_{\,\text{B}} = 5$$. Because $$4 < 5$$, the nuclear charge of beryllium is less than that of boron.

Now we consider the first ionisation enthalpy, commonly denoted $$\Delta_{i}H_{1}$$. This is the energy required to remove the outer-most electron from a gaseous atom in its ground state: $$ \text{X(g)} \;\longrightarrow\; \text{X}^{+}\!\text{(g)} + e^{-}, \quad \Delta_{i}H_{1}. $$ Across a period, the general trend is an increase in $$\Delta_{i}H_{1}$$ from left to right because the effective nuclear charge experienced by the valence electrons usually rises. However, there are well-known exceptions caused by the extra stability of completely filled or exactly half-filled subshells.

Beryllium has the electronic configuration $$ \text{Be}: 1s^{2}\,2s^{2}, $$ so its outer subshell $$2s$$ is completely filled. Boron, on the other hand, has $$ \text{B}: 1s^{2}\,2s^{2}\,2p^{1}, $$ leaving one electron in the higher-energy $$2p$$ subshell. Because a completely filled $$2s$$ subshell is relatively stable, removing an electron from Be requires more energy than removing the lone $$2p$$ electron from B.

Therefore, $$ \Delta_{i}H_{1}(\text{Be}) \; > \; \Delta_{i}H_{1}(\text{B}). $$ So, compared with boron, beryllium possesses a lesser nuclear charge but a greater first ionisation enthalpy.

Among the given alternatives, this description corresponds to Option B.

Hence, the correct answer is Option B.