= Solution
Start with <Taylor's theorem> with integral remainder, written on the whole interval using truncated powers:
$$
f(x)=\sum_{j=0}^k\frac{f^{(j)}(a)}{j!}(x-a)^j+\frac1{k!}\int_a^b(x-\theta)_+^k f^{(k+1)}(\theta)d\theta.
$$
For the <Peano kernel theorem>, require that $L$ be a linear error functional, annihilate all polynomials of degree at most $k$, and permit interchange with this integral. These conditions hold, in particular, for finite linear combinations of point evaluations of $f$ and its derivatives through order $k$, with the truncated-power derivatives interpreted almost everywhere in $\theta$; dominated integration justifies the interchange. Another sufficient setting is a bounded functional on a differentiability space where the truncated-power integral is convergent in its norm. Polynomial exactness alone, without linearity and an integration-interchange justification, is not sufficient.
Applying $L$ annihilates the Taylor polynomial and gives
$$
\boxed{L(f)=\frac1{k!}\int_a^bK(\theta)f^{(k+1)}(\theta)d\theta,\qquad K(\theta)=L_x[(x-\theta)_+^k].}
$$
This is the unnormalized <Peano kernel> convention used by the question; including $1/k!$ within the kernel instead removes the factor outside the integral.
For the stated derivative error, constants and linear and quadratic polynomials are annihilated, so $k=2$. Acting on $(x-\theta)_+^2$ gives
$$
K(\theta)=2(1-\theta)_+-\frac12(2-\theta)^2
=\begin{cases}-\theta^2/2,&0\le\theta\le1,\\-(2-\theta)^2/2,&1\le\theta\le2.\end{cases}
$$
Thus the <sharp central-secant derivative error> satisfies
$$
|L(f)|\le\frac12\int_0^2|K(\theta)|d\theta\,\|f'''\|_\infty
=\frac12\left(\frac16+\frac16\right)\|f'''\|_\infty.
$$
Hence $c=1/6$ works. It is the minimum: for $f(x)=x^3$, $L(f)=3-4=-1$ and $\|f'''\|_\infty=6$, so no smaller constant can satisfy the inequality. Therefore
$$
\boxed{c_{\min}=\frac16.}
$$
The sign-definite kernel explains why a constant third derivative attains the bound exactly.
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