I'm guessing you want your metric to be spherically symmetry and to tend asymptotically to flat spacetime. In that case you want something like:
$$ \mathrm ds^2 = -a(r)\mathrm dt^2 + \frac{\mathrm dr^2}{b(r)} + \mathrm d\Omega^2 $$
where both $a(r)$ and $b(r)$ have to tend to one for large $r$.
A $1/r$ force law is going to require that the Christoffel symbol $\Gamma^r_{tt}$ is approximately $1/r$. One quick thrash of Mathematica later and I get:
$$ \Gamma^r_{tt} = -\tfrac{1}{2}b(r)~\frac{\mathrm da(r)}{\mathrm dr} $$
As a quick check, for the Schwarzschild metric we expect $\Gamma^r_{tt}$ is approximately $1/r^2$ to give the inverse square law. For this metric:
$$ a(r) = b(r) = 1-\frac{2GM}{c^2r} $$
So:
$$ \Gamma^r_{tt} = -\left(1-\frac{2GM}{c^2r}\right)\frac{GM}{c^2r^2} $$
and in the limit of $r \rightarrow\infty$ we get $\Gamma^r_{tt}\propto 1/r^2$ as we expect. So far so good.
So you just need to find two functions $a(r)$ and $b(r)$ such that both tend to unity at large $r$ and:
$$ b(r)~\frac{\mathrm da(r)}{\mathrm dr} \approx \frac{1}{r}$$
for large $r$. Typically you'd look for functions like $1+f(r)$ where $f(r)$ becomes small at large $r$ and $\mathrm df/\mathrm dr \propto 1/r$, but that would give $f = \ln(r)$ and that doesn't go to unity at large $r$. No doubt our more experienced mathematicians can immediately think of a solution, but I have to confess that nothing springs to mind.
