Step-by-step Solution:

1. Identify the relevant physical principles: The decay of radioactive nuclei is governed by the law: $N(t) = N_0 e^{-\lambda t}$ where $N_0$ is the initial number of nuclei, $N(t)$ is the number of nuclei at time $t$, and $\lambda$ is the decay constant.

   The number of nuclei that have decayed by time $t$ is $\Delta N = N_0 - N(t)$. The fraction of nuclei that have decayed is given by: $f = \frac{\Delta N}{N_0} = \frac{N_0 - N(t)}{N_0} = 1 - \frac{N(t)}{N_0} = 1 - e^{-\lambda t}$

2. Calculate the decay constant ($\lambda$): The decay constant is related to the half-life ($T_{1/2}$) by the formula: $\lambda = \frac{\ln 2}{T_{1/2}}$ Given values are $T_{1/2} = 1386$ s and $\ln 2 = 0.693$. $\lambda = \frac{0.693}{1386} \text{ s}^{-1}$ We can simplify this calculation by noticing that $1386 = 2 \times 693$. $\lambda = \frac{0.693}{2 \times 693} = \frac{1}{2000} = 0.0005 \text{ s}^{-1}$

3. Calculate the fraction of decay in the given time: The time interval is $t = 80$ s. We need to calculate the value of $f = 1 - e^{-\lambda t}$. First, let's compute the exponent $\lambda t$: $\lambda t = (0.0005 \text{ s}^{-1}) \times (80 \text{ s}) = 0.04$

4. Evaluate the fraction: Now substitute the value of $\lambda t$ into the fraction formula: $f = 1 - e^{-0.04}$ Since the exponent $\lambda t = 0.04$ is very small compared to 1, we can use the Taylor series approximation $e^{-x} \approx 1 - x$ for $x \ll 1$. $f \approx 1 - (1 - \lambda t) = \lambda t$ $f \approx 0.04$ Note that the information about the initial activity ($10^3$ dps) is not needed to calculate the fraction of decayed nuclei.

5. Convert the fraction to percentage and round to the nearest integer: To express the fraction as a percentage, we multiply by 100: $\text{Percentage decay} = f \times 100\% \approx 0.04 \times 100\% = 4\%$

   For a more precise verification: Using a calculator, $e^{-0.04} \approx 0.96079$. $f = 1 - 0.96079 = 0.03921$. Percentage decay = $0.03921 \times 100\% = 3.921\%$. Rounding $3.921\%$ to the nearest integer gives $4\%$.

Final Answer:

The fraction of the initial number of nuclei that will decay in the first 80 s is 4%.