1. Let the decay constant of nucleus $n_1$ be $\lambda$.

   Then, for nucleus $n_2$: $\lambda_2 = 3\lambda_1 = 3\lambda$

2. Initially, the number of nuclei are same: $N_{10} = N_{20} = N_0$

3. We need the ratio after one half-life of $n_1$.

   The half-life of $n_1$ is: $T_{1/2,1} = \frac{\ln 2}{\lambda}$

4. Number of undecayed nuclei after time $t$ is: $N = N_0 e^{-\lambda t}$

5. For nucleus $n_1$, at time $t = T_{1/2,1}$: $N_1 = N_0 e^{-\lambda \cdot \frac{\ln 2}{\lambda}} = N_0 e^{-\ln 2} = \frac{N_0}{2}$

6. For nucleus $n_2$, at the same time $t = \frac{\ln 2}{\lambda}$: $N_2 = N_0 e^{-3\lambda \cdot \frac{\ln 2}{\lambda}} = N_0 e^{-3\ln 2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}$

7. Required ratio: $\frac{N_2}{N_1} = \frac{N_0/8}{N_0/2} = \frac{1}{4}$

8. Checking options:

   • A: $\frac{1}{4}$ ✅
   • B: $\frac{1}{8}$
   • C: $4$
   • D: $8$

Therefore, the correct answer is: $\boxed{\frac{1}{4}}$