A human body eliminates certain substances following a law...

The effective decay constant $\lambda_{\text{eff}}$ for a substance undergoing both biological excretion and radioactive decay is the sum of the biological decay constant $\lambda_b$ and the physical decay constant $\lambda_p$.

$\lambda_{\text{eff}} = \lambda_b + \lambda_p$

Since $\lambda = \dfrac{\ln 2}{T_{1/2}}$, the effective half-life $T_{\text{eff}}$ is given by:

$\dfrac{1}{T_{\text{eff}}} = \dfrac{1}{T_b} + \dfrac{1}{T_p}$

Given the biological half-life $T_b = 24 \text{ hours}$ and the physical half-life $T_p = 6 \text{ hours}$:

$\dfrac{1}{T_{\text{eff}}} = \dfrac{1}{24} + \dfrac{1}{6} = \dfrac{1 + 4}{24} = \dfrac{5}{24} \text{ hours}^{-1}$

$T_{\text{eff}} = \dfrac{24}{5} = 4.8 \text{ hours}$

The initial activity is $A_0 = 6 \mu\text{Ci}$ and the final activity is $A = 3 \mu\text{Ci}$. Since the activity reduces to exactly half of its initial value, the time elapsed is equal to one effective half-life.

$t = T_{\text{eff}} = 4.8 \text{ hours}$