= Solution
Ignoring interactions, every field has dimension $(d-2)/2$, and every Laplacian contributes two derivatives. The operator $\phi^n(\nabla^2\phi)^m$ therefore has dimension
$$
\Delta_{\mathcal O}=\frac{(n+m)(d-2)}2+2m.
$$
Because the integrated interaction $\int d^dx\,g\mathcal O$ is dimensionless, the coupling has scaling dimension
$$
\boxed{\Delta_g=d-\frac{(n+m)(d-2)}2-2m}.
$$
It is a <relevant coupling> when $\Delta_g>0$, a <marginal coupling> when $\Delta_g=0$, and an <irrelevant coupling> when $\Delta_g<0$.
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