**UPDATE: I have written a more complete answer here: How do the Einstein's equations come out of string theory?**

If you take the Polyakov (gravitons are bosons) Action:

$$S_P=-\frac{T}{2}\int \sqrt{\pm h}h^{\alpha\beta}\partial_\alpha X^\mu\partial_\beta X^\nu g_{\mu\nu}\mbox{ d}^2\xi$$

And take the gravitational terms of a somewhat "effective" spacetime action, you get

$$S_{G}=\lambda\int\left(R+\ell_s^2R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\right)\mbox{ d}^D x$$

Where we neglected terms of order $\ell_s^4$ and greater. Since $\ell_s$, the string length, is very small, this is approximately, (if we let ${\ell_s\rightarrow0}$)

$$S_{EH}=\lambda\int R\mbox{ d}^D x$$

The generalised (*n*-dimensional) EH Action. This is what is meant by string theory going to GR at the classical limit, the Polyakov Action simply goes down to the EH Action.

**Edit:** Also see JoshPhysics's answer here: In what limit does string theory reproduce general relativity?. The method I stated here is possible in principle, but is much more commplicated than the JoshPhysics's answer there. In his answer, he simply uses the Beta functional, $$\beta^G_{\mu\nu} = \ell_s^2 R_{\mu\nu}+\ell_s^4R_{\mu\nu}R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}+...$$ Then, setting the LHS to 0 to preserve conformal invariance:

$$R_{\mu\nu}+\ell_s^2R_{\mu\nu}R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}+...=0$$

For weak gravity, all terms except the first vanish, so that

$$R_{\mu\nu}=0$$