= Solution
Work in the supplied Planck units and write $t=T+\bar T>0$. The nonzero components of the <Kähler metric> and its inverse are
$$
K_{T\bar T}=\frac3{t^2},\qquad K_{C\bar C}=1,\qquad K^{T\bar T}=\frac{t^2}3,\qquad K^{C\bar C}=1.
$$
The <Kähler covariant derivatives of a superpotential> are
$$
D_TW=-\frac{3W}{t},\qquad D_CW=3C^2+\bar C(C^3+B).
$$
In the <supergravity F-term potential>, $K^{T\bar T}|D_TW|^2=3|W|^2$ cancels its negative universal term. This is the <no-scale supergravity> identity. The exact answer is therefore
$$
\boxed{V=\frac{e^{|C|^2}}{t^3}\left|3C^2+\bar C(C^3+B)\right|^2\ge0.}
$$
In particular $C=0$ is a global minimum for every $T$ in the domain, with \b[zero vacuum energy].
Use the standard convention for a <supergravity auxiliary field>, $F^i=-e^{K/2}K^{i\bar j}\overline{D_jW}$. In this convention
$$
F^T=e^{K/2}t\bar W,\qquad F^C=-e^{K/2}\left[3\bar C^2+C(\bar C^3+\bar B)\right].
$$
At the $C=0$ minimum these become
$$
\boxed{F^T=\frac{\bar B}{\sqrt t},\qquad F^C=0.}
$$
Thus <supersymmetry breaking> occurs if $B\ne0$, despite the vanishing vacuum energy. Indeed $K_{T\bar T}|F^T|^2=3|B|^2/t^3$ is precisely canceled by the negative <gravitino> contribution. \b[For $B=0$ the asserted breaking does not occur:] both <supergravity auxiliary fields> vanish at $C=0$, so this vacuum is supersymmetric.
For completeness, the full set of finite zero-energy minima includes additional branches. For $C=re^{i\theta}\ne0$, the condition $D_CW=0$ reduces, after factoring $re^{-i\theta}$, to
$$
r(3+r^2)e^{3i\theta}=-B.
$$
For every $B\ne0$ there is a unique positive solution of $r(3+r^2)=|B|$, since the left side increases strictly from zero to infinity. There are three associated phases,
$$
\boxed{r(3+r^2)=|B|,\qquad \theta=\frac{\arg(-B)+2\pi k}{3},\quad k=0,1,2.}
$$
These also give $V=0$, and on each branch $W=3B/(3+r^2)\ne0$. Consequently $F^C=0$ but $F^T\ne0$, so these are also <supersymmetry breaking> <no-scale vacua with a cubic matter superpotential>. A supersymmetric finite vacuum would require $D_TW=0$, hence $W=0$, and then $D_CW=3C^2=0$; this is possible only at $C=0$ with $B=0$.
The <scalar potential> is independent of $\operatorname{Im}T$ everywhere. On every zero-energy vacuum branch it is also independent of $\operatorname{Re}T$, so \b[both real components of $T$ are <flat directions of a scalar potential> at the vacuum]. Away from $D_CW=0$, the real direction is not flat: $\partial_tV=-3V/t$, giving a runaway towards $t\to\infty$ rather than a finite positive-energy stationary point. The zero-energy branches have no continuous <flat direction of a scalar potential> in $C$; their allowed $C$ values are isolated. This is why the vacuum does not select a unique numerical <gravitino> mass: the modulus $t$ is unfixed.
The <gravitino mass from a superpotential> is $m_{3/2}=e^{K/2}|W|$. On the conventional $C=0$ branch,
$$
\boxed{m_{3/2}=\frac{|B|}{(T+\bar T)^{3/2}},\qquad V_{\min}=0.}
$$
On a nonzero-$C$ branch it is instead
$$
\boxed{m_{3/2}=\frac{3|B|e^{r^2/2}}{(3+r^2)t^{3/2}}.}
$$
When $B=0$, the only finite zero-energy matter solution is $C=0$ and $m_{3/2}=0$, with unbroken <supersymmetry>. These results exhibit the distinction between vanishing vacuum energy and vanishing <supersymmetry breaking> <auxiliary fields> in a <no-scale supergravity> model.
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