Mass density of the wire in kg/m
3
$\rho = 8.5 \dfrac{\text{g}}{\text{cm}^3} \left( \dfrac{100 \, \text{cm}}{1 \, \text{m}} \right)^3 \left( \dfrac{1 \, \text{kg}}{1000 \, \text{g}} \right)$
$\rho = 8 500 \, \text{kg/m}^3$ answer
Volume of the wire
$M = \rho V$
$155(1000) = 8.5V$
$V = 18\,235.29 \, \text{ cc}$ answer
Cross-sectional area of the wire
$V = AL$
$18\,235.29 = A(1500 \times 100)$
$A = 0.121\,568\,627\,5 \, \text{cm}^2 (10 \, \text{mm} / \text{cm})^2$
$A = 12.16 \, \text{ mm}^2$ answer