[asy]
size(4cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);
[/asy]
Let $O$ be the center of the square. Since the side length of the square is $s$, the distance from $O$ to any vertex of the square is $\sqrt{s^2 + s^2} = s\sqrt{2}.$
Let $A,$ $B,$ $C,$ $D$ be the vertices of the square, in that order. Let $E,$ $F,$ $G,$ $H$ be the centers of the equilateral triangles with $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ $\overline{DA}$ as bases, respectively.
Let $X$ be the intersection of $\overline{EF}$ and $\overline{OG},$ and let $Y$ be the intersection of $\overline{GH}$ and $\overline{OF}.$
[asy]
size(6cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);
draw((0,0)--(1,-3^.5)--(2,0),dashed);
draw((0,2)--(-1,1)--(-3^.5,0)--(0,-1)--(0,0),dashed);
draw((2,0)--(3,1)--(2,2)--(1,1)--(1,0),dashed);
label("$A$",(0,2),NW);
label("$B$",(2,2),NE);
label("$C$",(2,0),SE);
label("$D$",(0,0),SW);
label("$E$",(1,-3^.5),S);
label("$F$",(2,0),E);
label("$G$",(-1,1),W);
label("$H$",(0,2),N);
label("$O$",(1,1),NW);
label("$X$",(5/3,-1/3),S);
label("$Y$",(4/3,4/3),E);
draw((1,1)--(5/3,-1/3),dashed);
draw((1,1)--(4/3,4/3),dashed);
[/asy]
Since $\triangle ABE$ is equilateral, $AE = BE = s.$ Since $O$ is the center of the square, $OE = OD - DE = s - \frac{s}{2} = \frac{s}{2}.$ Therefore, $OX = OE + EX = \frac{s}{2} + \frac{s\sqrt{3}}{2} = \frac{s(1+\sqrt{3})}{2}.$ Similarly, $OY = OE + EY = \frac{s}{2} + \frac{s\sqrt{3}}{2} = \frac{s(1+\sqrt{3})}{2}.$
Then, the side length of the large square is $XY = OX + OY = s(1 + \sqrt{3}).$
Therefore, the answer is $\boxed{s(1 + \sqrt{3})}.$
.
