In each of the two dimensions, each of the smaller squares has a range of 83 cm in which it can be located, with presumably a uniform distribution.  The centers, in each dimension, must be within 17 cm of each other in order to overlap.

Call the squares A and B. If square A is at, say, the leftmost possible position, square B has a range of 17 cm left-right to allow an overlap (I say allow, rather than cause, as the other dimension must also allow it, which I'll get to later). The same is true if square A is in the rightmost possible position.

But if square A is more than 17 cm from either left or rightmost position, there's a range of 34 cm for square B that allows an overlap. Since the range varies linearly with A's position, the weighted average of the positions of B that allow for overlap is (34*(17+34)/2 + 49*34)/83 = 2533/83 ~= 30.5180722891566.

The probability B will be in this range for that dimension is that divided by 83, or 2533/6889 or about 0.367687617941646.

Since the overlap must be in both dimensions, the needed probability is the square of that: 6416089/47458321 ~=  0.135194184387602.

A hundred million trials had an overap 13516884 times or 0.13516884 of the trials, which agrees.

succ=0;

for trial=1:100000000

  x1=rand*83;

  y1=rand*83;

  x2=rand*83;

  y2=rand*83;

  if abs(x1-x2)<17 && abs(y1-y2)<17

    succ=succ+1;

  end

end

disp(succ/trial)