= Solution
Treat the <orthonormal coframe in spacetime> and the Lorentz <connection 1-form> as independent variables, with nondegenerate coframe and $\omega_{ab}=-\omega_{ba}$. Fix the curvature convention
$$
R^a{}_b=d\omega^a{}_b+\omega^a{}_c\wedge\omega^c{}_b,qquad
T^a=dE^a+\omega^a{}_b\wedge E^b.
$$
The second expression is <Cartan's first structure equation>; the first is <Cartan's second structure equation>. Use compactly supported variations, or boundary data that make the variation's surface term vanish.
For a connection variation, $\delta R^{ab}=D\delta\omega^{ab}$, where $D$ is the <covariant exterior derivative>. Since $D\epsilon_{abc}=0$, its graded product rule gives
$$
\delta_\omega I=\int_M\epsilon_{abc}\delta\omega^{ab}\wedge T^c
+\int_{\partial M}\epsilon_{abc}\delta\omega^{ab}\wedge E^c.
$$
For a coframe variation, relabeling the dummy indices makes the three variations of the cubic term identical. Hence
$$
\delta_E I=\int_M\epsilon_{abc}\delta E^c\wedge
\left(R^{ab}+3\lambda E^a\wedge E^b\right).
$$
The independent variations therefore give $\epsilon_{abc}T^c=0$ and $\epsilon_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions the alternating symbol identifies an antisymmetric index pair with a single index, so these equations are equivalent to
$$
\boxed{T^a=0,\qquad R^{ab}=-3\lambda E^a\wedge E^b.}
$$
These are the equations of <first-order three-dimensional gravity>. The first selects the <Levi-Civita connection>; the second fixes <constant sectional curvature> $K=-3\lambda$. The factor of three follows from varying the coefficient exactly as printed.
On a static coordinate patch where $f>0$, choose the Lorentz frame metric $\eta_{ab}=\operatorname{diag}(-1,1,1)$ and coframe
$$
E^0=\sqrt f\,dt,\qquad E^1=\frac{dr}{\sqrt f},\qquad E^2=r\,d\theta.
$$
Put $A=f'/(2\sqrt f)$ and $B=\sqrt f/r$. Exterior differentiation gives $dE^0=-AE^0\wedge E^1$, $dE^1=0$, and $dE^2=BE^1\wedge E^2$. Solving the equation of vanishing <torsion forms> gives
$$
\omega^{01}=AE^0=\frac{f'}2dt,\qquad
\omega^{12}=-BE^2=-\sqrt f\,d\theta,\qquad
\omega^{02}=0.
$$
Here antisymmetry is in the two raised or two lowered frame indices; mixed time-space components have equal, rather than opposite, signs. This is the <Lorentzian connection-form antisymmetry> needed to retain the curvature signs.
Using <Cartan's second structure equation>, the nonzero <curvature 2-forms> are
$$
R^{01}=-\frac{f''}2 E^0\wedge E^1,\qquad
R^{02}=-\frac{f'}{2r}E^0\wedge E^2,\qquad
R^{12}=-\frac{f'}{2r}E^1\wedge E^2.
$$
For example, $d(AE^0)=-(\sqrt f A'+A^2)E^0\wedge E^1=-f''E^0\wedge E^1/2$, while the $02$ form is the product $\omega^0{}_1\wedge\omega^{12}=-AB E^0\wedge E^2$. Thus the field equations require $f''/2=3\lambda$ and $f'/(2r)=3\lambda$. Integrating the latter gives $f=C+3\lambda r^2$; the former is then satisfied automatically. The origin condition fixes $C=1$, yielding
$$
\boxed{f(r)=1+3\lambda r^2.}
$$
As a sign and normalization check, the <Ricci tensor> is $R_{\mu\nu}=-6\lambda g_{\mu\nu}$ and the <scalar curvature> is $R=-18\lambda$. The coframe calculation applies initially in the region with $f>0$; if $\lambda<0$, $f$ vanishes at $r=1/\sqrt{-3\lambda}$, delimiting that static chart without changing the derived function.
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