Mercury's average distance $D = 36.0$ million miles.

$T = 0.4D^{1.5} = 0.4(36)^{1.5}$

Since $36^{1.5} = 36^1 \cdot 36^{0.5} = 36\sqrt{36} = 36 \cdot 6 = 216$:

$T = 0.4(6^2)^{1.5} = 0.4(6^3) = 0.4 \cdot 216$

This matches choice G: $0.4(6^3)$.