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How can we see the Crab Nebula, when it is only 1,001 years old ...

How do astronomers calculate the distance of a star that is 5.74 million light years ... Answered 3 years ago · Author has 1.2K answers and 545.6K answer views ... and 1,001 years ago people on Earth saw it happening in real time from our frame ... Andre Engels, M.A. Mathematics & Physics, University of Groningen ( 1996).

For more information, see How can we see the Crab Nebula, when it is only 1,001 years old ...

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Space Math V

additional challenges in the math and physical science curriculum in grades 9 ... findings and compete for the necessary funding for their space craft. ... km/s the distance is (2500-136)/412 = 5.74 megaparsecs ( or 18,700,000 light ... x (1.737 x 108)3 = 2.2 x 1025 cm3, so the density = 7.4 x 1025 grams / 2.2 x 1025 cm3.

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Units & Measurement

number (or numerical measure) accompanied by a unit. Although the ... The base units for length, mass and time in these systems ... significant figures are 2, which is reliable and ... PHYSICS. 26. ◁ mass density is obtained by deviding mass by the volume of the substance. ... arithmetic operations can be understood from.

For more information, see Units & Measurement

Rounding off the Uncertain Digits

The preceding digit is left unchanged if the digit to be removed is less than 5

Example: 9.82 is rounded to 9.8.

The preceding digit is increased by one if the digit to be removed is greater than 5.

Example: 6.87 is rounded to 6.9

Read more: https://byjus.com/physics/error-significant-figures-rounding-off/

Given that

Substance mass m = 5.74 g

Volume of substance V = 1.2 cm^3

Formula of Density \rho = \frac{\text{mass}}{\text{volume}}

\rho = \frac{5.74}{1.2}

\rho = 4.78\bar{3}

Rounding to 4 significant digits

\rho = 4.8

Hence, density of the substance   \rho = 4.8 g cm^{-3}