Answer
After two half-lives, there will remain 20.0 mg of technetium-99m.
Work Step by Step
1. After a half-life, one-half of the technetium-99m decays. Therefore, the remaining mass is equal to $\frac{1}{2}$ of the initial mass.
Initial $^{99m}_{43}Tc$ mass: 80.0 mg
After one half-life:
$80.0$ $mg$ $\times \frac{1}{2} = 40.0$ $mg$
2. The same will occur on the second half-life, so, repeat the process:
After the second half-life:
$40.0$ $mg$ $\times \frac{1}{2} = 20.0$ $mg$