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).
