= Solution
Use the convention $\Theta^a{}_b=\frac12R^a{}_{bcd}e^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the <Ricci tensor> is diagonal. Contracting the six coefficients gives
$$
R_{00}=2B+F,\qquad R_{11}=R_{22}=R_{33}=-(2B+F).
$$
But $2B+F=2(1+q/2)+(1-q)=3$, so
$$
\boxed{R_{ab}=-3\eta_{ab},\qquad R=-12,\qquad G_{ab}=3\eta_{ab}.}
$$
In coordinate-independent form, $R_{\mu\nu}=-3g_{\mu\nu}$. Thus the vacuum <Einstein field equations> $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$ hold with
$$
\boxed{\Lambda=-3.}
$$
The curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual <curvature 2-forms> but cancels from their Ricci contraction. For $\alpha=0$ this is <Anti-de Sitter spacetime>; for nonzero $\alpha$ it is the <planar Einstein metric with cubic radial function>, with a nontrivial <Weyl tensor> rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.
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