We will work with the metric supplied,
$$\mathrm{d}s^2 = A(r)^2\mathrm{d}t^2 -B(r)^2 \mathrm{d}r^2 -r^2 \mathrm{d}\theta^2 - r^2 \sin^2\theta \, \mathrm{d}\phi^2$$
I will assume that omitting the fourth spatial coordinate (which I have added) is simply a typo in the original post. I have also redefined the arbitrary functions for convenience. We choose a basis,
$$e^{t} = A(r)\mathrm{d}t, \, \, \, e^{r}=B(r)\mathrm{d}r, \, \, \, e^{\theta} = r\mathrm{d}\theta, \, \, \, e^{\phi} = r\sin \theta \, \mathrm{d}\phi$$
We now compute the exterior derivative of all $e^{a}$, and re-express them in terms of the basis,
$$\mathrm{d}e^{t} = -\frac{A'}{AB} e^{t} \wedge e^{r}, \, \, \, \mathrm{d}e^{r} = 0, \, \, \, \mathrm{d}e^{\theta}=-\frac{1}{rB} e^{\theta} \wedge e^{r}, \, \, \, \mathrm{d}e^{\phi} =-\frac{1}{rB}e^{\phi}\wedge e^{r} - \frac{\cot \theta}{r} e^{\phi}\wedge e^{\theta}$$
The spin connection $\omega^{a}_{b}$ can be read off from Cartan's first equation,
$$\mathrm{d}e^{a} + \omega^{a}_{b}\wedge e^{b} = 0$$
As the exterior derivatives are already in terms of the basis, this is fairly straightforward:
$$\omega^t_r = \frac{A'}{AB} e^{t},\, \, \, \omega^{\theta}_{r} = \frac{1}{rB} e^{\theta}, \, \, \, \omega^{\phi}_{r} = \frac{1}{rB}e^{\phi}, \, \, \, \omega^{\phi}_{\theta} = \frac{\cot \theta}{r}e^{\phi}$$
All other connection components presumably vanish. To compute the curvature 2-form, $R^{a}_{b}$, the second Cartan equation is required,
$$R^{a}_{b} = \mathrm{d}\omega^{a}_{b} + \omega^{a}_{c}\wedge \omega^{c}_{b}$$
An example of a particular component,
$$R^{t}_{r}= -\left( \frac{A''B -A'B'}{AB^3}\right) e^{t}\wedge e^{r}$$
Often $\omega^{a}_{c}\wedge \omega^{c}_{b}$ will simply be zero because of the few non-zero spin connections. After you have found all the components of the 2-form, the Riemann tensor is given by,
$$R^{a}_{b} = \frac{1}{2}R^{a}_{bcd} e^{c}\wedge e^{d}$$
For our example, this implies,
$$R^{t}_{r t r} = -2\left( \frac{A''B -A'B'}{AB^3}\right)$$
Now recall that thus far we have been working in a particular orthonormal basis. To find the Riemann tensor in the coordinate basis, we use the relation,
$$R^{\lambda}_{\mu \nu \sigma}=(e^{-1})^{\lambda}_{a} R^{a}_{bcd} e^{b}_{\mu}e^{c}_{\nu}e^{d}_{\sigma}$$
where the l.h.s the curvature is in the coordinate basis, and $e^{a}_\mu \mathrm{d}x^\mu = e^{a}$. As you are aware, the Ricci tensor $R_{\mu \nu} = R^{\lambda}_{\mu \lambda \nu}$, which you can then use to solve $R_{\mu \nu} = 0$ which should, hopefully, be straightforward given this is a homework problem.
If you are unfamiliar with the Cartan formalism, see the gravitational physics lectures at http://perimeterscholars.org/, at approximately lecture 3-4.
$k$ is not a measure of the curvature of spacetime, but rather of the spatial sections. $k = 0$ doesn't mean that spacetime is flat (your example isn't: since $a(t) \neq a_0$, it can't be Minkowski spacetime), but rather that the spatial sections are just the Euclidean $\mathbb{R}^3$ space rather than, for example, a $3$-sphere.
It is also possible to get it the other way around: for $R = 0$, one can get a FLRW solution with $k \neq 0$. The Milne model is such an example (it is just the future light-cone of the origin in Minkowski spacetime).
As one can see from explicit computation for the general $k$ FLRW metric (or by looking up on Wikipedia, as I'm doing right now), the full expression for $R$ including the contribution from $k$ is
$$R = 6 \left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{k}{a^2}\right).$$
You can, though, compute the Ricci scalar for the spatial section. Using the metric
$$\text{d}s^2 = - \text{d}t^2 + a(t)^2 \left(\frac{\text{d}r^2}{1 - k r^2} + r^2 \text{d}\Omega^2\right),$$
which has spatial metric
$$\text{d}l^2 = \frac{a(t)^2 \text{d}r^2}{1 - k r^2} + a(t)^2 r^2 \text{d}\Omega^2,$$
I got (with the aid of the Mathematica package OGRE) the expression
$${}^{(3)}R = \frac{6k}{a^2(t)}.$$
Hence, you can understand $k$ as being proportional to the Ricci scalar of the spatial section.
Best Answer
The notion of asymptotic flatness in 4-dimensions was studied way back in 1962 by Bondi, van der Burg, Metzner (here) and Sachs (here) and more recently by Barnich and Troessaert (the first few papers here)
They described asymptotic flatness in terms of the Bondi coordinates, where the metric takes the form $$ ds^2 = \frac{V}{r} e^{2\beta} du^2 - 2 e^{2\beta} du dr + g_{AB} \left( dx^A - U^A du \right) \left( dx^B - U^B du \right) $$ where $A,B = 2,3$, $x^A = (\theta,\phi)$ and $\det g_{AB} = r^4 \sin^2\theta$. Bondi shows that every 4-dimensional metric can be written in the form above. Techniques developed later by Penrose showed that one should really set a more general condition wherein $\det g_{AB} = \frac{1}{4} r^4 e^{2 {\tilde\varphi}}$ (following Barnich's notation). An asymptotically flat spacetime then satisfies these boundary conditions (at large $r$)
$g_{AB} = r^2 \gamma_{AB} + {\cal O}(r)$ where $\gamma_{AB}$ is conformal to the metric of $S^2$, i.e. $$ \gamma_{AB}dx^A dx^B = e^{2\phi(u,\theta,\phi)} \left( d\theta^2 + \sin^2\theta d\phi^2 \right) $$
$\frac{V}{r} = - 2 r \partial_u {\tilde \varphi} + \Delta {\tilde \varphi} + {\cal O}(r^{-1})$
$\beta = {\cal O}(r^{-2})$
$U^A = {\cal O}(r^{-2})$
Given a metric, its asymptotic flatness can be checked by testing if the above boundary conditions hold.
EDIT: This is a discussion of asymptotic flatness at $\mathscr{I}^+$. An analogous discussion exists for $\mathscr{I}^-$ (simply take $u \to v = u + 2 r$. For a complete description of asymptotic flatness, one must also consider the structure of the spacetime at $i^0$. While this has been discussed in Ashtekar, Hansen, I do not know too much about it. I will leave this to other members to discuss.