The induced emf ($\varepsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:

$$\varepsilon = L \frac{dI}{dt}$$

where:

• $\varepsilon$ is the induced emf in the inductor,
• $L$ is the inductance of the inductor,
• $\frac{dI}{dt}$ is the rate of change of current through the inductor.

Given that the current $I = (3t + 8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.

$$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3$$

The given magnitude of induced emf is $12 \, \text{mV} = 12 \times 10^{-3} \, \text{V}$ (since $1\,\text{mV} = 10^{-3} \, \text{V}$).

Now, plug these values into the formula to find $L$:

$$12 \times 10^{-3} = L \cdot 3$$

Solving for $L$ gives:

$$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} = 4 \, \text{mH}$$

Therefore, the self-inductance of the inductor is 4 mH.