Is it the smallest?

As shown in the figure, the side length of the smallest square is $s = 99 + (99+17) \cos 45^\circ + 17 \approx 198.024 \implies \lfloor 100s \rfloor = \boxed{19802}$ .

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Proof: This is in response to @Thanos Petropoulos 's question. Referring to the figure above. The left side and the bottom side of the square are as tight fit as possible. Any variation in the square size comes from the right and top sides. Basically it depends on relative position of the small circle with the large circle. We have placed the small circle touching the big circle so as to be as tight as possible. Now the relative positions of the two circles only depend on the angle between the line joining the two circle centers and the $x$ -axis or any other reference line. Let take the $x$ -axis for convenient and call the angle $\theta$ . In the figure $\theta = 45^\circ$ . It can be seen that if $\theta < 45^\circ$ , the further right point of the small circle moves to the right or $x$ -direction hence increasing the size of the square. If $\theta > 45^\circ$ , it increases the $y$ -dimension and hence the size of the square. In fact the size of the square is symmetrical about $y=x$ or the $45^\circ$ -line. Anyway, I provide a mathematical proof here.

Consider $0^\circ \le \theta \le 45^\circ$ , all other ranges of $\theta$ are reflection of this range. Then the side length of the square is given by $a (\theta) = 99+116 \cos \theta + 17 = 116(1+\cos \theta$ . $a(\theta$ is smallest when $\cos \theta$ is smallest. For $0^\circ \le \theta \le 45^\circ$ , $\cos \theta$ is smallest when $\theta = 45^\circ$ . For $45^\circ \le \theta \le 90^\circ$ , $a(\theta) = 116(1+\red{\sin \theta}$ . Again $a(\theta)$ is smallest when $\theta = 45^\circ$ .

For the tightest fit, consider a square as in the picture, with the smaller circle tangent to two adjacent sides of the square, the larger circle tangent to the two opposite sides, and the two circles tangent to each other. The distance between the two centers is $17 + 99 = 116$ .

To find the width of this square, we add the lengths of the two radii to the side length of the right isosceles triangle whose hypotenuse connects the two circles' centers. With a hypotenuse length of $116$ , the two legs each have length $\frac{116}{\sqrt{2}} = 82.024$ . Thus the side length of the square is $116 + 82.024 = 198.024$ .

Note that $198$ is the minimal side length possible for a square containing a circle of radius $99$ . What I've calculated is the horizontal distance from the left side of the larger circle to the right side of the smaller circle. Had this quantity been less than $198$ , the answer would have been $198$ .