For the two-point graph, $L=2$ and $I=2$, so its <superficial degree of divergence> in three dimensions is
$$
D=3L-2I=2.
$$
Its additive mass-squared counterterm is therefore proportional to $\lambda\Lambda^2$; writing that counterterm as $m^2\delta_m$ gives the dimensionless scaling $\delta_m\sim\lambda\Lambda^2/m^2$, or $\Lambda^2/m^2$ when coupling factors are suppressed as in the question. For the six-point graph, $L=2$ and $I=3$, so $D=0$ and its ultraviolet divergence is logarithmic:
$$
\delta_\lambda\sim\lambda^2\log(\Lambda/m).
$$
Suppressing powers of $\lambda$ yields the two stated estimates.

Solved by gpt-5.6-sol high.

Solution

Paper 304 / 2 / b / ii / Solution

{title2=$D$}
{wiki}

The superficial degree of divergence $D$ is the ultraviolet power predicted by rescaling all independent loop momenta together. For a scalar graph with $L$ loops and $I$ propagators behaving as $p^{-2}$ in $d$ dimensions, $D=dL-2I$ before cancellations and subdivergences are considered.


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