= Solution
Write $E(h)$ for the supplied approximation. The <asymptotic expansion> is
$$
Y=E(h)+ch^2+r(h),\qquad r(h)=O(h^3),
$$
where $O$ is <Big O notation>. At the refined parameter $h/3$,
$$
Y=E(h/3)+\frac{ch^2}{9}+r(h/3).
$$
Multiply this identity by nine and subtract the first one. The unknown leading <truncation error> coefficient cancels:
$$
8Y=9E(h/3)-E(h)+9r(h/3)-r(h).
$$
Thus the <Richardson extrapolation> estimate is
$$
Y_{\rm ext}(h)=\frac{9E(h/3)-E(h)}8,\qquad
Y-Y_{\rm ext}(h)=O(h^3).
$$
The two supplied values give
$$
\boxed{Y_{\rm ext}(1)=\frac{9\cdot3-2}{8}=\frac{25}{8}=3.125.}
$$
The correction to the finer approximation is $(3-2)/8=1/8$. Neglecting the higher-order remainder would also give $c=9/8$.
The conclusion is asymptotic: the leading error order improves from two to at least three when both parameter values are in the range of the stated <asymptotic expansion>. With no bound on the remainder constant, the two finite observations alone do not guarantee that $25/8$ is closer to $Y$. Indeed, for any proposed value $L$, the <polynomial> model
$$
E(h)=L-c_Lh^2-d_Lh^3,\qquad
c_L=13L-\frac{79}{2},\qquad d_L=\frac{75}{2}-12L
$$
satisfies $E(1)=2$, $E(1/3)=3$ and the required <asymptotic expansion> with true <limit> $L$. This is why <finite observations do not certify Richardson extrapolation accuracy>: asymptotic cancellation provides an improved-order estimator, while a numerical error certificate needs quantitative remainder information.
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