In Birrell and Davies, the author says in the Introduction that
If the gravitational field is treated as a small perturbation, and attempts are made to quantize it along the lines of quantum electrodynamics (Q.E.D.), then the square of the Planck length appears in the role of coupling constant.
But why does the square of the Planck length comes in as a coupling constant here? How does one determine what the coupling constant should be?
Let's for simplicity work in units where $\hbar=1=c$. If we try to quantize perturbatively the Einstein-Hilbert (EH) action $$ S~=~ \frac{1}{16\pi G} \int \! d^dx \sqrt{-g} (R-2\Lambda)\tag{1} $$ in $d$ spacetime dimensions, we see that the coupling constant $G$ has dimension $$[G] ~=~ L^{d-2},\tag{2}$$ since $$[S]~=~ L^0, \quad [x^{\mu}]~=~L^1, \quad [g_{\mu\nu}]~=~L^0, \quad [R]~=~L^{-2}~=~[\Lambda].\tag{3}$$
In fact the Planck length $\ell_P$ is defined as the combination of the 3 fundamental physics constants $c$, $\hbar$ and $G$ that has dimension of length, so it is not surprising that $G$ is proportional to $\ell_P^{d-2}$, cf. the statement in Birrell & Davies.
For more information, see e.g. my related Phys.SE answers here and here.
