= Solution
The <Lagrange interpolation polynomials> at $c_1=1/3$ and $c_2=1$ are
$$
\ell_1(s)=\frac32(1-s),\qquad \ell_2(s)=\frac32s-\frac12.
$$
Their <integrals> give
$$
\left(\int_0^{c_i}\ell_j(s)\,ds\right)_{ij}
=\begin{pmatrix}5/12&-1/12\\3/4&1/4\end{pmatrix},
\qquad
\left(\int_0^1\ell_j(s)\,ds\right)_j=(3/4,1/4).
$$
Thus the coefficients are exactly those of a <collocation Runge-Kutta method>. More explicitly, with stage slopes $F_j=f(t_n+c_jh,Y_j)$, the degree-two <polynomial>
$$
p(t_n+sh)=y_n+h\sum_{j=1}^2F_j\int_0^s\ell_j(r)\,dr
$$
satisfies $p(t_n)=y_n$, $p(t_n+c_ih)=Y_i$, and $p'(t_n+c_ih)=F_i$. These are the <collocation Runge-Kutta method> equations, and $y_{n+1}=p(t_n+h)$. For an <implicit Runge-Kutta method>, the nearby stages exist uniquely for sufficiently small $h$ under <Lipschitz continuity>; see <stage solvability of an implicit Runge-Kutta method>.
Write $e=(1,1)^T$, $c=Ae$, and interpret powers of $c$ componentwise. The <Butcher order conditions> through degree three are verified directly:
$$
b^Te=1,\qquad b^Tc=\frac12,\qquad b^Tc^2=\frac13,\qquad b^TAc=\frac16.
$$
For the last equality, $Ac=(1/18,1/2)^T$. However, the necessary degree-four <Butcher order condition> fails:
$$
b^Tc^3=\frac{5}{18}\ne\frac14.
$$
Consequently, for sufficiently smooth <ordinary differential equations>, the <local truncation error> is $O(h^4)$ and the <order of a Runge-Kutta method> is exactly
$$
\boxed{p=3}.
$$
This is the two-stage <Radau IIA method>; the conclusion follows from the explicit calculations rather than its name.
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