Let's work through the problem step by step.

The decay of a radioactive nucleus is given by the formula:

$$N(t) = N_0 \, e^{-\lambda t},$$

where:

$N(t)$ is the number of nuclei remaining at time $t$,

$N_0$ is the initial number of nuclei,

$\lambda$ is the decay constant.

For nucleus $n_1$ with decay constant $\lambda_1$, its half-life $t_{1/2}$ is:

$$t_{1/2} = \frac{\ln 2}{\lambda_1}.$$

After one half-life of $n_1$, the number of $n_1$ nuclei remaining is:

$$N_1 = N_0 \, e^{-\lambda_1 \, t_{1/2}} = N_0 \, e^{-\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-\ln2} = \frac{N_0}{2}.$$

Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$:

$$\lambda_2 = 3\lambda_1.$$

For nucleus $n_2$, the number of nuclei remaining after one half-life of $n_1$ (which is $t = \frac{\ln2}{\lambda_1}$) is:

$ N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}. $

Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time:

$ \frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1/8}{1/2} = \frac{1}{4}. $

Therefore, the correct ratio is $\frac{1}{4}$, which corresponds to Option A.