= Solution
Neglecting the $\epsilon_V$ contribution to the tilt, the observed value gives
$$
-\frac1{30}\simeq-\frac{2M_{\rm Pl}^2m^2}{\Lambda^4},
\qquad
\frac{M_{\rm Pl}^2m^2}{\Lambda^4}\simeq\frac1{60}.
$$
Part (d) then gives $\phi_{60}=e^{-1}\phi_e$. Using the formal end estimate from part (b), $\epsilon_V(\phi_{60})\simeq e^{-2}$. The amplitude relation therefore yields
$$
\frac{H}{M_{\rm Pl}}
=\sqrt{8\pi^2A_s\epsilon_V(\phi_{60})}
\simeq\frac{\sqrt{8\pi^2(2\times10^{-9})}}{e}
\simeq1.5\times10^{-4}.
$$
Thus the requested order-of-magnitude estimate is
$$
\boxed{H/M_{\rm Pl}\sim10^{-4}}.
$$
The simultaneous approximations make $\epsilon_V(\phi_{60})$ only moderately small, so this numerical result should be read as the order-of-magnitude estimate requested rather than a precision fit.
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