The answer is #950cm^3# of water must be added.
Let's start with the initial solution. We know that we have an initial volume of #V_1 = 300.0cm^3# and a density of #rho = 1.38 g/(cm^3)#. Automatically, we can find out what the mass of the solution is by using
#rho = m_(solution)/V_1 -> m_(solution) = rho * V_1 = 1.38 g/(cm^3) * 300.0 cm^3 = 414g#
We know that we're dealing with a #48%# percent concentration by mass solution of #H_2SO_4#, which means that we have #48g# of #H_2SO_4# for every #100g# of solution. This means that the mass of sufuric acid, which represents the solute is equal to
#c = m_(solute)/m_(solution) * 100%#
#m_(solute) = (c * m_(solution))/(100%) = (48% * 414g)/(100%) = 198g#
The number of moles of sulfuric acid, knowing its molar mass of #98.0g/(mol)#, is equal to
#n_(H_2SO_4) = m/(molarmass) = (198g)/(98.0g/(mol)) = 2.02# moles
The acid's molarity, expressed in moles per #dm^3#, is equal to
#C_1 = n/V = (2.02mol es)/(300.0 * 10^(-3) dm^3) = 6.73(mol es)/(dm^3)#
We can now use a simple dilution calculation to determine the volume of the desired solution
#C_1 * V_1 = C_2 * V_2 -> V_2 = C_1/C_2 * V_1#
#V_2 = 6.73/1.61 * 300.0 cm^3 = 1250 cm^3#
This means that the volume of water that needs to be added is #1250 - 300 = 950cm^3#.
You can check the answer by determining the second solution's density
#rho_2 = m_(solution2)/V_2 = (414g + 950g)/(1250cm^3) approx 1.10 g/(cm^3)# - what is given for the second solution.
