The liquid CHBr3 has a density of 2.89 g dm-³. What volume of this liquid should be measured to contain a total of 4.8×10²⁴ molecules of CHBr3 ?

Hello!
The molar mass of C H Br_3 (bromoform) is about
12 + 1 + 3*80 = 253 (g/(mol)).
Therefore one mole of this substance has a mass of about 253 g.
One mole of any substance contains N_A approx 6*10^(23) molecules (this constant is called Avogadro's constant). Hence the given number of molecules represents  (4.8*10^(24))/(6*10^(23)) = 8 (moles), and from the above paragraph their mass is about 8*253 = 2024 (g).
Finally, volume may be computed as mass divided by the density, because rho = m/V. In this case it is
V = (2024 g)/(2.89 g/((cm)^3)) approx 700 (cm)^3.
This is the same as 0.7 dm^3.
Note that the density of C H Br_3 is incorrectly stated in the question as 2.89 g/((dm)^3). Actually it is 2.89 g/((cm)^3). This liquid is much more dense than water (about 1 g/(cm^3) ).
https://pubchem.ncbi.nlm.nih.gov/compound/Bromoform