= Solution
Assume $L$ is linear, annihilates all <polynomials> of degree at most $k$, is defined on the truncated-power kernels and permits interchange with the remainder integral. For $k\geq1$, a convenient sufficient condition is that $L$ be a <bounded linear functional> on $C[a,b]$; more general approximation functionals need the corresponding continuity/interchange property on their domain. Taylor's integral remainder gives
$$
f(x)=\sum_{j=0}^k\frac{f^{(j)}(a)}{j!}(x-a)^j
+\frac1{k!}\int_a^b(x-\theta)_+^k f^{(k+1)}(\theta)\,d\theta.
$$
Applying $L$, the <polynomial> terms vanish, and the <Peano kernel theorem> gives
$$
\boxed{L(f)=\frac1{k!}\int_a^bK(\theta)f^{(k+1)}(\theta)\,d\theta,\qquad
K(\theta)=L[(x-\theta)_+^k].}
$$
Here $t_+=\max(t,0)$. This is the printed raw-kernel convention; the normalized <Peano kernel> includes $1/k!$ inside its definition instead, and then has no factorial outside the integral. The interchange is justified by boundedness applied to the limit of Riemann sums of the continuous kernel-valued integrand. For a kernel with $k=0$, its discontinuity requires a function space/interchange assumption accommodating the step functions as well.
For the specified <Simpson's rule> error, direct evaluation on $1,x,x^2,x^3$ gives zero, so $k=3$. For $0\leq\theta\leq2$,
$$
K(\theta)=\frac{(2-\theta)^4}{4}
-\frac43(1-\theta)_+^3-\frac13(2-\theta)^3.
$$
Simplifying separately on the two halves gives
$$
\boxed{K(\theta)=\begin{cases}
-\theta^3(4-3\theta)/12,&0\leq\theta\leq1,\\
-(2-\theta)^3(3\theta-2)/12,&1\leq\theta\leq2.
\end{cases}}
$$
Both pieces are strictly negative in the open interval; they agree at the midpoint. Their integral is $-1/15$, as direct <polynomial> integration shows. Thus
$$
|L(f)|\leq\frac1{6}\int_0^2|K(\theta)|\,d\theta\,\|f^{(4)}\|_\infty
=\boxed{\frac1{90}\|f^{(4)}\|_\infty.}
$$
This is optimal, not merely an upper bound. For $f(x)=x^4/24$, the fourth <derivative> is identically one and $L(f)=-1/90$, so \b[the minimum constant is $\rho=1/90$].
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