= Solution
The condition $2m=\hbar^2$ makes
$$
H=-\frac{d^2}{dx^2}-V_0e^{-x^2}.
$$
For
$$
\psi_a(x)=e^{-ax^2/2},
\qquad a>0,
$$
the squared norm is the <Gaussian integral>
$$
\langle\psi_a|\psi_a\rangle
=\int_{\mathbb R}e^{-ax^2}\,dx
=\sqrt{\frac\pi a}.
$$
Since $\psi_a'=-ax\psi_a$, integration by parts gives the kinetic quotient
$$
\frac{\int\psi_a(-\psi_a'')\,dx}
{\int\psi_a^2\,dx}
=\frac{\int|\psi_a'|^2\,dx}{\int\psi_a^2\,dx}
=a^2\frac{\int x^2e^{-ax^2}\,dx}
{\int e^{-ax^2}\,dx}
=\frac a2.
$$
The potential quotient is
$$
-V_0
\frac{\int e^{-(a+1)x^2}\,dx}
{\int e^{-ax^2}\,dx}
=-V_0\sqrt{\frac a{1+a}}.
$$
The variational principle therefore yields the <Gaussian variational bound for an attractive Gaussian well>
$$
\boxed{
E_0\leq E(a)
=\frac a2-V_0\frac{\sqrt a}{\sqrt{1+a}}}.
$$
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