Each side of a metallic cube of mass 5.580 kg is measured to be 9.0...

Mass of the metallic cube, $m = 5.580$ kg (4 significant figures)

Side length of the cube, $a = 9.0$ cm $= 9.0 \times 10^{-2}$ m (2 significant figures)

Volume of the cube, $V = a^3 = (9.0 \times 10^{-2})^3 = 729 \times 10^{-6}$ m$^3$

Density of the material, $\rho = \dfrac{m}{V} = \dfrac{5.580}{729 \times 10^{-6}} = 7.6543... \times 10^3$ kg m$^{-3}$

In multiplication or division, the final result should retain as many significant figures as are present in the original number with the least significant figures. Here, the side length ($9.0$ cm) has the least number of significant figures, which is 2.

Rounding off $7.6543... \times 10^3$ to 2 significant figures, we look at the third digit. Since the third digit is $5$ and is followed by non-zero digits, the preceding digit is increased by $1$.

$\rho = 7.7 \times 10^3$ kg m$^{-3}$

Comparing this with $X \times 10^3$ kg m$^{-3}$, we get $X = 7.7$.

Answer: $7.7$