jmc

We have that $$Q(\sqrt{3})=a_0+a_1\sqrt{3}+3a_2+3a_3\sqrt{3}+\cdots =20+17\sqrt{3},$$so $$\begin{array}{rl}{a}_0+3a_2+9a_4+81a_6+\cdots & =20,\\ a_1+3a_3+9a_5+81a_7+\cdots & =17.\end{array}$$Since $0\le a_i<3,$ the problem reduces to expressing 20 and 17 in base 3. Since $20=2\cdot 9+0\cdot 3+2$ and $17=9+2\cdot 3+2,$ $$Q(x)=x^5+2x^4+2x^3+2x+2.$$In particular, $Q(2)=\boxed{86}.$