Since the bugs start out walking perpendicularly, and there is nothing in the problem to alter this symmetry, the bugs are always walking perpendicularly. Since each bug is walking perpendicularly to the line separating it from the bug chasing it, the gap is closing at the speed of the chasing bug. Therefore, each bug walks a distance equal to the side of the square before it meets the next bug.
In order to conveniently express the equation of the bugs' motion, use standard polar coordinates, and let the first bug's position at any instant be (r(t), theta(t)). Assume the initial square is centered at the origin.
Then by symmetry the bugs are always at four corners of some square centered at the origin, and if they meet they meet at the origin. Also, each bug is always walking in a direction pi/4 (45 degrees) away from the radial line to the origin. This means that in the limit as the time step goes to zero, the bug travels sec(pi/4) = sqrt(2) units along its path for every unit of progress made good toward the center. Since the corners are initially d/sqrt(2) distance from the center, each bug travels distance d before they meet, assuming they meet at all.
Since a bug's path always crosses radial lines at angle psi = pi/4, the path is a logarithmic spiral with angle psi = pi/4 and equation
Moreover since the bugs walk clockwise, both r(t) and theta(t) decrease as t increases, in other words r increases as theta increases, hence a is positive. Also, psi = pi/4 gives us
(this is easiest to see by drawing the path for a small time step delta-t and taking the limit as delta-t goes to 0). The solution is
(that is, a = 1). As we know, this spiral makes infinitely many "wraps" around the origin as the radius approaches zero, but it does have finite length from any point inward and its limit point is the origin, where the bugs will meet (unless one wants to quibble about the behavior at the exact limit point).
How much closer is the bug to the origin after its first complete cycle around the origin? Recall that r(0) = d/sqrt(2). As the bug walks clockwise around the origin, after one full circuit its angle decreases from theta(0) to theta(t1), where t1 (the time at which full circuit occurs) is defined by
r(0) = e^(theta(0) + b) r(t1) = e^(theta(0) - 2*pi + b)
The quantity we want is