The answer is approximately #1041# #"g/L"#.
Since you start with a concentration of 1M, let's assume you have 1 L of solution. That means that you have
#M = n/V => n = M * V = ("1 mole")/("L") * 1L = 1 "mole"#
This means that you have a mass of #NaCl# equal to
#m_(NaCl) = n * "molar mass" = 1 "mole" * 58.5 "g"/("mol") = 58.5# #"g"#
In order to determine its density, you must calculate the percent concentration by mass of the #NaCl# solution. This is done by dividing the mass of the solute (#NaCl#) by the total mass of the solution, and multiplying by 100%. Water is assumed to have a density of #"1 g/mL"# (1 L of water = 1000 grams)
#%"mass" = m_(NaCl)/m_("sol") * 100%= (58.5g)/(58.5 + 1000g) * 100%#
#%"mass" = 5.53%#
You'd then turn to a density table like this one:
http://www.maelabs.ucsd.edu/mae171/Conc%20vs%20density.pdf
According to this, a #5.53# percent concentration by mass corresponds to approximately #1041# #"g/L"#. (Notice that the mass of #NaCl# at that density is 62.478 g, relatively close to 58.5 g, your value).
