See/z/q168138057.htm for the answer.

2

Let the height S-ABC of the pyramid be SH, H be the outer center (inner heart, center of gravity) of the regular triangle ABC, and the connected AH and BC intersect at m,

Link SM, AM=√3BC/2=(√3/2)*2√6=3√2, HM=AM/3=√2,

sm=√(mh^2+sh^2)=√3,s△sbc=sm*bc/2=√3*2√6/2=3√2，

S△ABC=√3/4(2√6)^2=6√3，

VS-ABC=S△ABC*SH/3=2√3，

VS-ABC =(S△ABC+S△SBC+S△SAC+S△SAB)* R/3 =(6√3+3 * 3√2)R/3，

(6√3+3*3√2)R/3=2√3，

∴R=√6-2.