A steel wire 1.6 mm across breaks at a stress of 4.8 × 10⁸ Pa.

Step 1. Area. $A = \pi(0.80 \times 10^{-3}\ \text{m})^2 = 2.01 \times 10^{-6}\ \text{m}^2$.

Step 2. Breaking load. $4.8 \times 10^{8}\ \text{Pa} \times 2.01 \times 10^{-6}\ \text{m}^2 = 965\ \text{N}$, the weight of $\dfrac{965\ \text{N}}{9.8\ \text{m/s}^2} = 98.5\ \text{kg}$.

Step 3. Safe load. A fifth: $193\ \text{N}$, or $19.7\ \text{kg}$.