From a neutral atom where a electron is removed then it become cation and when a electron is added to the atom then it becomes anion.

For a atom $$x$$ when it loose a electron then it becomes cation $$x - {e^ - } \to {x^ + }$$

Let atomic no. of $$x$$ is $$z$$

then no. of proton in $${x^ + } = z$$

and no. of electron in $${x^ + } = z - 1$$

So the ratio $${z \over e}$$ as no. of electrons decreases. And when no of electron decreases then per electron attraction increases from nucleus and radius of cation deceases.

Similarly when atom $$x$$ becomes $${x^ - }$$ by adding a electron $$\left( {x + {e^ - } \to {x^ - }} \right)$$ then no. of proton in $${x^ - } = z$$ and no. of electron in $${x^ - } = z + 1.$$

Now the ratio of $${z \over e}\,\,$$ in $${x^ - }$$ decreases as no. of electron increases. As electron increases in $${x^ - }$$ then attraction from nucleus decreases on per electron and the distance between electron and nucleus increases so the radius also increase in anion.

All $$4$$ ions are from second period and for $${O^{2 - }}\,\,$$ and $${F^ - }$$ both have $$1s,$$ $$2s$$ and $$2p$$ shell and for $$L{i^ + }\,\,$$ and $${B^{3 + }}$$ both have $$1s$$ shell.

So, size of $${O^{2 - }}\,\,\,$$ and $$\,\,\,{F^ - }\,\,\,$$ are more then $$L{i^ + }\,\,\,\,$$

and $$\,\,\,\,{B^{3 + }}$$ and among $${O^{2 - }}\,\,\,$$ and $$\,\,\,\,{F^ - }$$

For $$\,\,\,$$ $${O^{2 - }},\,\,{z \over e} = {8 \over {10}} = 0.8$$

For $$\,\,\,$$ $${F^ - },\,\,\,{z \over e} = {9 \over {10}} = 0.9$$

as for $${O^{2 - }},\,\,{z \over e}\,\,\,$$ is less so per electron attraction from nucleus decreases and radius increases more than $${F^ - }.$$