 Solution

The tableau as printed is not a collocation tableau. Its second stage row is $(3/4,3/4)$, whose sum is $3/2$, while the corresponding printed node is $1$. This violates the collocation tableau row-sum identity. The original PDF confirms this coefficient; it is not only a TeX transcription defect.

The required integrated basis at nodes $1/3,1$ isso direct integration givesThus replacing the last stage coefficient by $1/4$ gives the intended two-stage Radau IIA collocation method. Its quadrature moments areIt integrates quadratics exactly but not cubics, so the collocation order theorem gives order three. Independently, $b^{T}Ac=1/6$, confirming the remaining third-order Butcher order condition.

For completeness the literal printed Runge-Kutta method has only order one. For the autonomous equation $y^{\prime }=y$, the Taylor expansion of its stability function beginswhereas $e^z=1+z+z^2/2+O(z^3)$. Since $b^{T}1=1$, it is consistent, but this discrepancy already rules out order two. Therefore neither the requested equivalence nor the third-order conclusion is true without the stated coefficient repair.