= Solution
For fixed $n$, the right-hand side of the quantization equation tends to zero with $\varepsilon$. If
$$
\pi\varepsilon\left(n+\frac12\right)<f(1)=\frac{16}{3},
$$
the unique energy satisfies $E_n<1$, so the first branch of the <action integral for a linear-to-square-root potential> applies exactly. Solving
$$
\frac{16}{3}E_n^{3/2}
=\pi\varepsilon\left(n+\frac12\right)
$$
gives
$$
\boxed{
E_n=\left[
\frac{3\pi\varepsilon}{16}
\left(n+\frac12\right)
\right]^{2/3}.}
$$
Thus this formula is valid whenever
$$
\boxed{
\varepsilon<\frac{16}{3\pi(n+1/2)}.}
$$
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