A complex problem

First, remember that $|a + bi| = \sqrt{a^2 + b^2}$ . Note that without the conditions $0 < a < 10, -10 < b < 10$ , the solution set for $\sqrt{a^2 + b^2} \leq 7$ is the disk $a^2 + b^2 \leq 49$ , which is centered at $(a, b) = (0,0)$ with radius $r = 7.$ Keeping this in mind and restricting $a$ and $b$ to $0 < a < 10, -10 < b < 10$ , the probability that $\sqrt{a^2 + b^2} \leq 7$ is the area of the solution set divided by the area of all the possible values of $a$ and $b$ . This is $x = \frac{\frac{\pi r^2}{2}}{\Delta a \Delta b} = \frac{(\frac{49\pi}{2})}{200} = \frac{49\pi}{400}$ Next, find that $\frac{1}{x^3} = \frac{64000000}{117649\pi^3} \approx 17.54$ . The solution is therefore $\lfloor\frac{1}{x^3}\rfloor = \boxed{17}$