How many tons of lead is needed to curve space 1 nanometer? How many tons of lead is  needed to curve space 1 nanometer? 
 A: If you use this answer and its references, you should be able to estimate this easily.
In the notation of that answer, if you measure out a length $\mathrm{d} R$ radially with a measuring rod of length $\mathrm{d} R$, the increase in radial co-ordinate $r$ is given by:
$$\mathrm{d} R = \frac{\mathrm{d} r}{\sqrt{1-\frac{2\,G\,\mathcal{M}(r)}{c^2\,r}}}$$
where $\mathcal{M}(r)$ is the mass contained inside a sphere defined by the radial co-ordinate $r$. In the case of a constant density $\rho$ sphere, the discrepancy between $r$ and a total $R$ measured by you by laying out measuring rods end to end is given by (I'm using the binomial theorem to approximate):
$$R\approx \int_0^r (1+\frac{4}{3\,c^2}\, G\, \pi\, \rho\, {r^\prime}^2)\,\mathrm{d}\,r^\prime$$
so that the discrepancy is estimated by:
$$R-r \approx \frac{4}{9\,c^2}\, G\, \pi\, \rho\, r^3 = \frac{1}{3}\,\frac{G\,M}{c^2}$$
or one sixth the Schwarzschild radius for a mass $M$, meaning you will need quite a few elephants, or tonnes of lead, to see the discrepancy of the magnitude you seek.
You can also glean the same formula from Feynman's elementary introduction to GR in chapter 42 of the second volume of the lectures (equation 42.3). Incidentally, the discrepancy works out to about 1.5mm for the Earth at its surface.
