= Solution
Use the classical dilute-gas convention
$$
\frac{p}{nk_BT}=1+B_2(T)n+O(n^2),
$$
so the <second virial coefficient> has units of <volume (thermodynamics)>. The <Mayer f-function> formula for an isotropic <pair potential> is
$$
B_2(T)=2\pi\int_0^\infty[1-e^{-\beta\phi(r)}]r^2\,dr.
$$
Only separations smaller than $a$ contribute. Define $x=\beta\varepsilon>0$ and substitute $t=r^3/a^3$, so $r^2dr=(a^3/3)dt$:
$$
B_2(T)=\frac{2\pi a^3}{3}\int_0^1[1-e^{-x(1-t)}]dt
=\boxed{\frac{2\pi a^3}{3}\left[1-\frac{1-e^{-x}}x\right],\qquad x=\frac{\varepsilon}{k_BT}.}
$$
This is the <second virial coefficient of a bounded cubic repulsion>. For every finite positive $T$, $1-e^{-x}<x$, so $B_2(T)>0$. There is no finite positive <Boyle temperature>. At high <temperature>, expanding the exponential gives
$$
B_2(T)=\frac{2\pi a^3}{3}\left(\frac x2-\frac{x^2}{6}+O(x^3)\right)
\sim\frac{\pi a^3\varepsilon}{3k_BT}\longrightarrow0^+.
$$
Thus the intended limiting statement is
$$
\boxed{T_B=+\infty\quad\text{in the sense that }B_2(T)\to0\text{ only as }T\to\infty.}
$$
At the other extreme the classical expression tends to $2\pi a^3/3$, the hard-exclusion value for separation $a$. This is a check on the <integral>, not a claim that a real molecular gas stays classical at arbitrarily low <temperature>. A molar <virial expansion> would instead use $B_{2,\mathrm m}=N_A B_2$.
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