= Solution
The largest field in each gap lies at that gap's smaller radius. Thus the quantity to minimize is
$$
E_{\max}(V_2)=\max\left\{\frac{c|V_1-V_2|}{a(c-a)},\frac{b|V_2|}{c(b-c)}\right\}.
$$
For positive $V_1$, projecting any $V_2$ outside $[0,V_1]$ back to that interval reduces both absolute potential drops, so the optimum lies inside it. Within the interval, one candidate maximum decreases linearly with $V_2$ and the other increases linearly. Their maximum is minimized at their intersection, not by making the potential drops equal.
With $a=0.1\ \mathrm m$, $c=0.2\ \mathrm m$, $b=0.4\ \mathrm m$ and $V_1=1000\ \mathrm V$, the two maxima are $20(1000-V_2)$ and $10V_2$ in $\mathrm{V\,m^{-1}}$. Equating them gives
$$
\boxed{V_2=\frac{2000}{3}\ \mathrm V\simeq666.7\ \mathrm V,\qquad E_{\max}=\frac{20000}{3}\ \mathrm{V\,m^{-1}}\simeq6667\ \mathrm{V\,m^{-1}}}.
$$
The maximum is halved relative to the two-sphere system and is attained at both $a^+$ and $c^+$. At $c^-$ and $b^-$ the optimized field is only about $1667\ \mathrm{V\,m^{-1}}$; at $c$ the charged middle shell produces a jump.
For plotting, the two-sphere <electric potential> is $V^{(2)}(r)=V_1(1/r-1/b)/(1/a-1/b)$. In the three-sphere system it is
$$
V^{(3)}(r)=\begin{cases}
V_2+(V_1-V_2)\dfrac{1/r-1/c}{1/a-1/c},&a\le r\le c,\\
V_2\dfrac{1/r-1/b}{1/c-1/b},&c\le r\le b.
\end{cases}
$$
The <electric potential> remains continuous at $c$, while its radial derivative changes there. The graphs show both the continuous potentials and the one-sided gap fields; the marked endpoints identify all maxima.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-concentric-sphere-field-grading.png]
{title=Potential and electric field before and after optimized intermediate-sphere bias}
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