 Solution

Substitution gives $\dot{\varphi }=2M_{\mathrm{Pl}}/t$, $\ddot{\varphi }=-2M_{\mathrm{Pl}}/t^2$ and $V=V_{\ast }{e}^{-{\varphi }_{\ast }/M_{\mathrm{Pl}}}(t_{\ast }/t{)}^2=10M_{\mathrm{Pl}}^2/t^2$, exactly satisfying both equations when ${\varphi }_{\ast }=M_{\mathrm{Pl}}\log (V_{\ast }{t}_{\ast }^2/(10M_{\mathrm{Pl}}^2))$. Since $a\propto t^2$, $\ddot{a}>0$: it inflates. $N=2\log (t_{\ast }/t)$ and $\Delta \varphi =2M_{\mathrm{Pl}}\log (t_{\ast }/t)=N{M}_{\mathrm{Pl}}$, so 60 e-folds require a field excursion about $60M_{\mathrm{Pl}}$.