Two solid bodies together?

Since the entire disc is outside the radius of the sphere, we can treat the sphere as a point particle. Since the sphere is inside the sphere of radius $3a$ that contains the disc, we cannot treat the disc as a point particle. The gravitational attraction between the sphere and the disc will be on a line between the two centres, though. Thinking about a circular disc of radius $r$ inside the disc, the force of gravitational attraction between the two objects is $\begin{aligned} \int_0^{3a} \frac{G M \frac{2\pi r \,dr}{9\pi a^2} m}{16a^2 + r^2} \times \frac{4a}{\sqrt{16a^2 + r^2}} & = \; \frac{8GMm}{9a}\int_0^{3a} \frac{r\,dr}{(16a^2 + r^2)^{\frac32}} \\ & = \; \frac{8GMm}{9a}\Big[ - \frac{1}{\sqrt{16a^2 + r^2}}\Big]_0^{3a} \\ & = \; \frac{8GMm}{9a} \left(\frac{1}{4a} - \frac{1}{5a}\right) \; = \; \frac{2GMm}{45a^2} \end{aligned}$ making the answer $2 + 45 = \boxed{47}$ .