Answer
$$7.3 \times 10^{11} \space kg/m^3$$
Work Step by Step
In order to calculate the density of the black hole, we have to find its mass and its volume. 1. Mass $$mass_{black \space hole} = 1 \times 10^3 \space mass_{sun}$$ To calculate the mass of the sun, we will need its volume and its density. $r_{sum} = 7.0 \times 10^5 \space km \times \frac{1000 \space m}{1 \space km} = 7.0 \times 10^8 \space m$ $$V_{sphere} = \frac 43 \pi r^3 = \frac 43 \pi (7.0 \times 10^8 \space m)^3 = 1.4 \times 10^{27} \space m^3$$ $$1.4 \times 10^{27} \space m^3 \times \frac{1.4 \times 10^3 \space kg}{m^3} = 2.0 \times 10^{30} \space kg$$ $$mass_{black \space hole} = 1 \times 10^3 ( 2.0 \times 10^{30} \space kg) = 2.0 \times 10^{33} \space kg$$ 2. Volume $$r_{black \space hole} = \frac 12 r_{moon}$$ $$r_{moon} = \frac 12 d_{moon} = \frac 12 2.16 \times 10^3 \space miles \times \frac{1609 \space m}{1 \space mile} = 1737720 \space m$$ $$r_{black \space hole} = \frac 12 (1737720 \space m) = 868860 \space m$$ $$V_{black \space hole} = \frac 43 \pi (868860 \space m)^3 = 2.75 \times 10^{18} \space m^3$$ 3. Density $$\rho = \frac{m}{V} = \frac{2.0 \times 10^{30} \space kg}{2.75 \times 10^{18} \space m^3} = 7.3 \times 10^{11} \space kg/m^3$$
