= Solution
Inserting an independently biased conducting sphere makes two vacuum gaps, each with its own constant radial <electric flux>. Let $A_1$ and $A_2$ be the coefficients of $1/r^2$. The two potential drops are
$$
V_1-V_2=A_1\left(\frac1a-\frac1c\right),\qquad V_2=A_2\left(\frac1c-\frac1b\right).
$$
Therefore
$$
\boxed{\mathbf E(r)=\begin{cases}
\dfrac{(V_1-V_2)ac}{(c-a)r^2}\hat{\mathbf r},&a<r<c,\\
\dfrac{V_2bc}{(b-c)r^2}\hat{\mathbf r},&c<r<b.
\end{cases}}
$$
The middle conductor can carry net <electric charge>, so $A_1$ and $A_2$ need not be equal. Its charge is $4\pi\epsilon_0(A_2-A_1)$, consistent with the jump in the normal <electric field>. For $0<V_2<V_1$ both fields point outwards; outside that interval one of the gap fields reverses direction.
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