= Solution
Put $I_0=\sqrt{2\pi/m^2}$ and assume $m^2>0$. The normalized free <Gaussian integral> has covariance $\Delta=1/m^2$. Expanding the interaction exponential and using <Wick's theorem> gives the <Feynman rules>: a quartic vertex contributes $-\lambda$, each internal <propagator> contributes $\Delta$, and each <vacuum diagram> is divided by its <Feynman-diagram symmetry factor>. Multiply the resulting sum by $I_0$. Because we are expanding $I$, rather than $\log I$, disconnected <vacuum diagrams> must also be included.
At order $\lambda^2$ there are two quartic vertices and four internal <propagators>. Let $r$ be the number of edges joining the two vertices. The remaining half-edges must pair at their own vertex, so $r$ is even. The possibilities $r=0,2,4$ exhaust the <vacuum diagrams>:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-48-vacuum-diagrams.png]
{title=All three second-order quartic vacuum diagrams and their symmetry factors}
For $r=0$, each vertex has three possible pairings, giving $N_0=9$ <Wick contractions>. For $r=2$, choose the two half-edges at each vertex, then pair the chosen half-edges across the vertices: $N_2=\binom42^2 2!=72$. The remaining two half-edges at each vertex form a self-loop. For $r=4$, all half-edges pair across the vertices, giving $N_4=4!=24$. The common denominator from the expansion is $2!(4!)^2=1152$, so the <Feynman-diagram symmetry factors> are
$$
S_0=\frac{1152}{9}=128,\qquad S_2=\frac{1152}{72}=16,\qquad S_4=\frac{1152}{24}=48.
$$
Each <vacuum diagram> has the same factor $I_0\lambda^2\Delta^4$ before division by its <Feynman-diagram symmetry factor>. Consequently the \b[coefficient of $\lambda^2$] is
$$
\boxed{\frac{I_0}{m^8}\left(\frac1{128}+\frac1{16}+\frac1{48}\right)=\frac{35}{384m^8}\sqrt{\frac{2\pi}{m^2}}.}
$$
As an independent check, the eighth moment obtained from the <Gaussian moment pairing theorem> is $\langle x^8\rangle_0=7!!/m^8=105/m^8$, and $105/[2!(4!)^2]=35/384$. The first terms of the <zero-dimensional quartic perturbation series> are therefore
$$
\frac{I}{I_0}=1-\frac{\lambda}{8m^4}+\frac{35\lambda^2}{384m^8}+O(\lambda^3).
$$
Strictly, the printed phrase “Taylor expansion” means a formal expansion or a right-sided <asymptotic expansion>. The general coefficient has absolute successive ratio $(4v+3)(4v+1)/[24(v+1)m^4]\to\infty$, so the <power series> has zero radius of convergence. Nevertheless, for $\lambda\geq0$ the remainder after any fixed truncation of $e^{-\lambda x^4/4!}$ is bounded by the next absolute term, and its <Gaussian integral> is finite. This proves the displayed <asymptotic expansion> as $\lambda\downarrow0$. For negative real $\lambda$ the defining integral diverges.
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