= Solution
Divide the <primordial tensor power spectrum> by the <slow-roll curvature power spectrum>:
$$
r=\frac{8H^2/(4\pi^2M_{\rm pl}^2)}{H^4/(4\pi^2\dot\phi^2)}=\frac{8\dot\phi^2}{M_{\rm pl}^2H^2}.
$$
Differentiating the scalar-field <Friedmann equation> and using the <inflaton> equation gives $\dot H=-\dot\phi^2/(2M_{\rm pl}^2)$. Hence $r=16\epsilon_H\simeq16\epsilon$. For the quadratic model at 50 remaining e-folds,
$$
\boxed{r\simeq\frac{16}{101}\simeq0.158.}
$$
This is above the supplied upper bound, \b[so the model fails the stated tensor constraint], despite its satisfactory scalar tilt and adjustable scalar amplitude. Calibrating $m$ does not reduce $r$, since the coupling cancels from the ratio.
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