jmc

The base has area $B=4\cdot 4=16.$ The length of the segment connecting a vertex of the base to the center of the base equals $\sqrt{2^2+2^2}=2\sqrt{2}.$ This segment and the altitude of the pyramid form a right triangle with hypotenuse of length $4$. Thus we have $h=\sqrt{4^2-(2\sqrt{2}{)}^2}=\sqrt{8}$. By the volume formula for a pyramid, we have $V=Bh/3=16\sqrt{8}/3\approx \boxed{15.08}$ cubic inches.