Step-by-step Solution:

1. Understanding the relationship between activity and half-life. The activity $A$ of a radioactive sample at any time $t$ is given by the formula: $A(t) = A_0 e^{-\lambda t}$ where $A_0$ is the initial activity at $t=0$, and $\lambda$ is the decay constant.

   The half-life, $T_{1/2}$, is the time it takes for the activity to reduce to half its initial value. It is related to the decay constant by $T_{1/2} = \frac{\ln 2}{\lambda}$.

   We can express the activity after a time $t$ in terms of the number of half-lives, $n = \frac{t}{T_{1/2}}$. $A(t) = A_0 e^{-(\frac{\ln 2}{T_{1/2}})t} = A_0 e^{-\ln 2 \cdot (t/T_{1/2})} = A_0 (e^{\ln 2})^{-t/T_{1/2}} = A_0 (2)^{-t/T_{1/2}}$ $A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} = A_0 \left(\frac{1}{2}\right)^n$ This formula states that after $n$ half-lives, the activity is $A_0$ divided by $2^n$.

2. Analyzing the given information. We have two radioactive sources, $S_1$ and $S_2$.

   • At $t=0$, their activities are the same. Let's denote this initial activity as $A_0$. So, $A_{01} = A_{02} = A_0$.
   • We need to find the ratio of their activities, $A_1/A_2$, at a later time.
   • The activity $A_1$ is measured when source $S_1$ has completed its $3^{\text{rd}}$ half-life. This means for $S_1$, the number of half-lives elapsed is $n_1 = 3$.
   • The activity $A_2$ is measured when source $S_2$ has completed its $7^{\text{th}}$ half-life. This means for $S_2$, the number of half-lives elapsed is $n_2 = 7$.

3. Calculating the activity $A_1$ for source $S_1$. Using the formula from Step 1, with $n_1 = 3$ and initial activity $A_0$: $A_1 = A_{01} \left(\frac{1}{2}\right)^{n_1} = A_0 \left(\frac{1}{2}\right)^3 = \frac{A_0}{8}$

4. Calculating the activity $A_2$ for source $S_2$. Similarly, for source $S_2$, with $n_2 = 7$ and initial activity $A_0$: $A_2 = A_{02} \left(\frac{1}{2}\right)^{n_2} = A_0 \left(\frac{1}{2}\right)^7 = \frac{A_0}{128}$

5. Calculating the ratio $A_1 / A_2$. Now we can find the required ratio: $\frac{A_1}{A_2} = \frac{A_0/8}{A_0/128}$ The initial activity $A_0$ cancels out: $\frac{A_1}{A_2} = \frac{1/8}{1/128} = \frac{128}{8}$ Simplifying the fraction: $\frac{A_1}{A_2} = 16$

   Alternatively, $\frac{A_1}{A_2} = \frac{(1/2)^3}{(1/2)^7} = \left(\frac{1}{2}\right)^{3-7} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16$

Final Answer:

The ratio $A_1 / A_2$ is 16.