1. **Problem statement:** We have defects occurring along a cable following a Poisson process with rate $\lambda = 0.2$ defects per kilometre. Given no defects in the first 2 kilometres, we want the conditional probability of no defects between kilometre 2 and 3. 2. **Poisson process property:** The number of defects in disjoint intervals are independent. So, the number of defects between 2 and 3 km is independent of the number in the first 2 km. 3. **Distribution:** The number of defects in an interval of length $t$ follows a Poisson distribution with parameter $\lambda t$. 4. **Calculate probability:** For the interval from 2 to 3 km, length $t=1$, the parameter is $\lambda t = 0.2 \times 1 = 0.2$. 5. **Probability of no defects:** The probability of zero defects in a Poisson($\mu$) distribution is $P(X=0) = e^{-\mu}$. 6. **Apply formula:** $$ P(\text{no defects between 2 and 3}) = e^{-0.2} \approx 0.8187 $$ 7. **Interpretation:** Since the process is memoryless and independent in disjoint intervals, the conditional probability given no defects in the first 2 km is the same as the unconditional probability for the interval 2 to 3 km. **Final answer:** Approximately $0.82$, which corresponds to option c.