= Solution
A dimensionless <action> in six spacetime dimensions requires the <Lagrangian density> to have <mass dimension> six. The <kinetic term> then fixes $2+2[\phi]=6$, and each <field interaction term> must have the same <mass dimension>. Therefore
$$
\boxed{[\phi]=2,\qquad [g_1]=0,\qquad [g_2]=-2.}
$$
Equivalently, in powers of a length unit $L$, $\phi$ has units $L^{-2}$, $g_1$ has units $L^0$, and $g_2$ has units $L^2$. Thus literal \i[length exponents] would be $(-2,0,2)$; the $d$'s in the supplied <superficial degree of divergence> formula must instead mean <mass dimensions>, namely $(2,0,-2)$. This distinction resolves the wording's convention mismatch.
To check the convention independently, let $I,V_3,V_4,\ell$ count internal lines, cubic <interaction vertices>, quartic <interaction vertices>, and independent <loop momenta>. A connected <Feynman diagram> obeys $2I+n=3V_3+4V_4$ and $\ell=I-V_3-V_4+1$. Hence its <superficial degree of divergence> is $6\ell-2I=6-2n+2V_4$, agreeing with the supplied formula only with the <mass dimensions> above.
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