Question 1026227
Look at either circle near one of the corners of the square.  Draw a radius which intersects a side of the square; which is also tangent to the circle.  This tangency point, and the center point of the circle, and the corner of the square, form a right isosceles triangle having two sides of 1 unit.  The hypotenuse can be found:


h for this hypotenuse, {{{h=sqrt(1^2+1^2)=sqrt(2)}}}.


You can find the length of the diagonal of the square.  The diagonal contains SIX of the radii lengths (of any of the circles); but also understand the distance from corner to a nearest center of a circle...
and revise the sum of lengths which compose the diagonal of the outer square.


{{{sqrt(2)+r+2r+r+sqrt(2)}}}-----the length of the diagonal of the square, which simplified is...
{{{highlight_green(4r+2sqrt(2))}}}.


If this solution is stopped here, can you finish answering the question?