= Solution
Write the <reduced Planck mass> as $M=(8\pi G)^{-1/2}=m_{\rm pl}/\sqrt{8\pi}$, and use natural units. Near the flat maximum, the <quartic hilltop inflation> trajectory has $V\simeq\alpha\sigma^4$ and $V'\simeq-12\alpha\phi^3$. The <slow-roll approximation> gives $H^2\simeq V/(3M^2)$ and $3H\dot\phi\simeq-V'$. Therefore the <number of e-folds> is
$$
N(\phi)\simeq\frac1{M^2}\int_\phi^{\phi_{\rm upper}}\frac{V}{-V'}d\phi
\simeq\frac{\sigma^4}{24M^2}\left(\frac1{\phi^2}-\frac1{\phi_{\rm upper}^2}\right).
$$
The lower field value dominates this integral. Thus
$$
\boxed{\phi_{50}^2\simeq\frac{\sigma^4}{1200M^2}
=\frac{\pi\sigma^4}{150m_{\rm pl}^2},\qquad
\phi_{50}\simeq0.145\frac{\sigma^2}{m_{\rm pl}}.}
$$
This is far below $\sigma$ for the assumed small symmetry-breaking scale. Strict slow roll actually fails before the minimum: $|M^2V''/V|\sim1$ at $\phi^2\sim\sigma^4/(36M^2)$. The last approach to the minimum is not described by slow roll. Using that earlier upper limit changes the dominant fifty-e-fold estimate by an order-one <number of e-folds>, not its leading scale dependence. No extrapolation of slow roll all the way to the minimum is needed.
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