Is a hydrogen molecule smaller than hydrogen atom?
You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen:


  Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids.


From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is eerily similar to $4 \times \pu{0.53 Å}= \pu{2.12 Å}$.

However this is a distraction from the main point of the statement in the book, which is simply that concentration of electron density between the two $\ce{H}$ nuclei results in an "end-to-end" distance (the distance, using some measure to determine the edges of the molecule, along the internuclear axis) that is less$^\dagger$ than $\pu{2.12 Å}$, and that the internuclear distance is less than $\pu{2\times\pu{0.53 Å}= 1.06 Å}$ (which it is: the H-H bond length is $\pu{0.74 Å}$). An internuclear distance shorter than twice the lone H radius is consistent with bonding, a process by which electrons and nuclei are allowed to occupy lower-energy regions of the potential than they would when the atoms are far appart.

But the answer to the question is no, a hydrogen molecule is still bigger than a hydrogen atom. 



$\dagger$ But just barely. Define the edge of an atom or molecule as the point where the electron density equals that at a distance of $\pu{1 Bohr}$ ($\pu{0.529 Å}$) from the nucleus in a hydrogen atom. Next find the edges in a simple  MO model of $\ce{H2}$ ([3]). The edges are $\pu{1.056 Å}$ from the center of the molecule, that is, the molecule along the longest axis has a diameter of $\pu{2.11 Å}$, barely less than $4\times$ the Bohr radius. 



[3]: Pearson, Ralph G., and Palke, William E.. Proc.Natl.Acad.Sci.USA, Vol. 77, No.4, pp.1725-1727, April 1980.