= Solution
The bare coupling does not depend on the <renormalization scale>. Expressing it through the renormalized field and coupling gives, to the stated order,
$$
0=\mu\frac d{d\mu}
\left[\mu^\epsilon(g+\delta g)-2\mu^\epsilon g\delta Z\right].
$$
At one loop $\delta Z=0$. Put $a=3/(16\pi^2)$ and $\delta g=ag^2/\epsilon$. If $\beta_g=\mu\,dg/d\mu$, differentiation and expansion through $g^2$ yield
$$
\boxed{\beta_g=-\epsilon g+\frac{3g^2}{16\pi^2}+O(g^3)}.
$$
Since $\lambda=\mu^\epsilon g$,
$$
\beta_\lambda=\mu\frac{d\lambda}{d\mu}
=\epsilon\lambda+\mu^\epsilon\beta_g
=\frac{3\mu^{-\epsilon}\lambda^2}{16\pi^2}+O(\lambda^3),
$$
and in four dimensions
$$
\boxed{\beta_\lambda=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3)}.
$$
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