= Solution
Under the <slow-roll approximation>,
$$
H^2\simeq\frac{8\pi G}{3c^2}V,
\qquad
3H\dot\phi\simeq-V'(\phi).
$$
For $V=m^2\phi^2/2$ and the branch $\phi>0$, set
$$
A=\sqrt{\frac{4\pi G}{3c^2}}.
$$
Then $H\simeq Am\phi$ and
$$
\dot\phi\simeq-\frac{m}{3A}
=-\frac{mc}{\sqrt{12\pi G}}.
$$
With $\phi(0)=\phi_0$ and $a(0)=a_0$,
$$
\boxed{\phi(t)=\phi_0-\frac{mc}{\sqrt{12\pi G}}t},
$$
and integrating $\dot a/a=H$ gives
$$
\boxed{a(t)=a_0\exp\left[\frac{2\pi G}{c^2}
\left(\phi_0^2-\phi(t)^2\right)\right]}.
$$
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