Answer:
[tex]\large \boxed{0.300\text{ lb/yd}^{3}}[/tex]
Explanation:
[tex]\rho = \dfrac{m}{V}\\\\m = \text{0.890 g } \times \dfrac{\text{1 lb}}{\text{453.6 g}} = 1.962 \times 10^{-3}\text{ lb}\\\\V = \text{5.00 L} \times \dfrac{\text{1 yd}^{3}}{\text{764.6 L}} =6.539 \times 10^{-3} \text{ yd}^{3}\\\\\rho = \dfrac{1.962 \times 10^{-3}\text{ lb}}{6.539 \times 10^{-3} \text{ yd}^{3}} = \textbf{0.300 lb/yd}^{3}\\\\\text{The density of helium is }\large \boxed{\mathbf{0.300}\textbf{ lb/yd}^{\mathbf{3}}}[/tex]
