A transmission cable of length 18 km produces noise of 5 x 10-10 W. The power output is 5 x 10-8 W. Calculate the minimum signal-to-noise ratio and attenuation, if the attenuation per unit length is 4 dBkm-1.

Signal power ($P_S$) $= 5 \times 10^{-8}$ W  
Noise power ($P_N$) $= 5 \times 10^{-10}$ W

$$
SNR = \frac{P_S}{P_N} = \frac{5 \times 10^{-8} \text{ W}}{5 \times 10^{-10} \text{ W}} = 100
$$

$$
SNR_{dB} = 10 \log_{10}(SNR) = 10 \log_{10}(100) = 10 \times 2 = 20 \text{ dB}
$$
The minimum signal-to-noise ratio is $\boxed{20 \text{ dB}}$.

Length ($L$) $= 18$ km  
Attenuation per unit length ($\alpha$) $= 4 \text{ dBkm}^{-1}$

$$
Attenuation = \alpha \times L = 4 \text{ dBkm}^{-1} \times 18 \text{ km} = 72 \text{ dB}
$$
The total attenuation is $\boxed{72 \text{ dB}}$.