= Solution
For the quadratic potential, the e-fold integral above gives $\Phi_N/M_{\rm Pl}\simeq\sqrt{4N+2}$. At $N\simeq60$, the field is $\Phi_*/M_{\rm Pl}\simeq\sqrt{242}\simeq15.6$. Since $k=aH$ at exit and $N$ counts down,
$$
\frac{d\ln k}{dN}=-1+\epsilon_H,\qquad
\frac{d(\Phi/M_{\rm Pl})}{dN}=\frac2{\sqrt{4N+2}}.
$$
To leading slow-roll order, $|\Delta N|\simeq\Delta\ln k$. The <quadratic-inflation field interval from a wavenumber interval> is therefore
$$
\boxed{\frac{|\Delta\Phi|}{M_{\rm Pl}}
\simeq\frac2{\sqrt{242}}\,\Delta\ln k
\simeq0.64\quad\text{using }\Delta\ln k\simeq5.}
$$
The corresponding field endpoints are approximately $\sqrt{232}\,M_{\rm Pl}$ and $\sqrt{252}\,M_{\rm Pl}$, or $15.2$ to $15.9$ reduced Planck masses. Longer wavelengths exit earlier at the larger field value. Using $\ln100=4.605$ instead of the supplied rounded logarithmic interval gives a span of about $0.59M_{\rm Pl}$; retaining $1-\epsilon_H$ changes either estimate by less than one percent. The observable interval samples a small part of the large background field value, rather than the entire excursion to the end of inflation.
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