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You want a cold disk.

Mercury is an example of a volatile, which, for our purposes, means that it exists as a solid only at very low temperatures. One thesis (Funk (2015)) classifies it as "moderately volatile" (Table 1.1), also noting that its 50% condensation temperature, $T_{c}$¹, is less than that of sulfur. It is, in fact, around 250 K. That is really, really, really low. For a graphical comparison, see Figure 2 of Albarede (2009).

Therefore, abundances of Mercury in the center of the protoplanetary disk were very low (Evolution of Earth and its Climate, page 64). The thing is, that's where terrestrial planets form - inside the frost line (though there should be regions inside the frost line where mercury can still condense). Sure, there's some mercury there - after all, we have it on Earth, and some of that likely formed with the planet - but not nearly enough to create an entire planetary core.

Therefore, you need a really cold planetary disk. HL Tauri appears to be a possibility. Carrasco-Gonzalez et al. (2016) found a temperature of $\sim70\text{-}140\text{ K}$ at 10 AU. Given that temperatures in disks roughly obey a power law² of $$T(r)=T(r_o)\left(\frac{r}{r_o}\right)^{-q}$$ and $q=0.5\text{-}0.6$, then if we take the lower temperature bound, we find that $T(1\text{ AU})$, for instance, can be $221\text{ K}$, which would seem to work. There still might not be enough mercury for a whole core to form, but one can hope.

¹ $T_c$ is the temperature at which 50% of all the mercury is condensed.
² In general, disks around Sun-like stars follow a power law of $$T(r)\propto\left(\frac{r}{1\text{ AU}}\right)^{-q}\left(\frac{M_*}{M_{\odot}}\right)$$ where $M_*$ is the mass of the star and $q>0$. The proportionality constant is probably somewhere in the range of $200\text{-}300\text{ K}$; these slides give it as $280\text{ K}$.

posted over 7 years ago