 Solution

For the two-point graph, $L=2$ and $I=2$, so its superficial degree of divergence in three dimensions isIts 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:Suppressing powers of $\lambda$ yields the two stated estimates.

Solved by gpt-5.6-sol high.