= Solution
The nodes $0,1/2,1$ have <Lagrange interpolation polynomials>
$$
\ell_1(s)=2s^2-3s+1,\quad
\ell_2(s)=4s-4s^2,\quad
\ell_3(s)=2s^2-s.
$$
The coefficients obtained from $a_{ij}=\int_0^{c_i}\ell_j(s)\,ds$ and $b_j=\int_0^1\ell_j(s)\,ds$ reproduce the given <Butcher tableau>. In particular, the second row's last entry is $-1/24$, which is missing from the TeX transcription. Thus this is the three-stage <Lobatto IIIA method>.
One applicable collocation theorem is that $s$-stage Lobatto IIIA collocation has order $2s-2$. Alternatively, the <fourth-order conditions for a Runge-Kutta method> give a direct verification. With $e=(1,1,1)^T$, $c=Ae$ and $C=\operatorname{diag}(c)$, the eight required <Butcher order conditions> are
$$
\begin{gathered}
b^Te=1,\quad b^Tc=\tfrac12,\quad b^Tc^2=\tfrac13,\quad b^TAc=\tfrac16,\\
b^Tc^3=\tfrac14,\quad b^TCAc=\tfrac18,\quad
b^TAc^2=\tfrac1{12},\quad b^TA^2c=\tfrac1{24}.
\end{gathered}
$$
Powers of $c$ here are componentwise. Substitution satisfies all eight. The <stability function> computed from the stages is
$$
R(z)=1+zb^T(I-zA)^{-1}e
=\frac{1+z/2+z^2/12}{1-z/2+z^2/12}.
$$
Its expansion satisfies $R(z)-e^z=-z^5/720+O(z^6)$, so even the scalar linear problem fails the fifth-order condition. Hence
$$
\boxed{p=4.}
$$
The <Butcher order condition> theorem equates order $p$ with all conditions for <rooted trees> through order $p$; the displayed conditions are its complete specialization through order four.
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