 Solution

Write the reduced Planck mass as $M=(8\pi G{)}^{-1/2}=m_{\mathrm{pl}}/\sqrt{8\pi }$, and use natural units. Near the flat maximum, the quartic hilltop inflation trajectory has $V\simeq \alpha {\sigma }^4$ and $V^{\prime }\simeq -12\alpha {\varphi }^3$. The slow-roll approximation gives $H^2\simeq V/(3M^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$. Therefore the number of e-folds isThe lower field value dominates this integral. ThusThis is far below $\sigma$ for the assumed small symmetry-breaking scale. Strict slow roll actually fails before the minimum: $|M^2{V}^{\prime \prime }/V|\sim 1$ at ${\varphi }^2\sim {\sigma }^4/(36M^2)$. 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.