Titration question- help please

A student performed an investigation to determine an accurate concentration of phosphoric acid (H3PO4) in rust remover. 10mL of the rust remover was diluted to 200mL in a volumetric flask. A 20mL aliquot of the diluted acid was then titrated against a standardised 0.930 mol/L sodium hydroxide solution. The volume of sodium hydroxide required to reach the endpoint was 21.3mL.

Calculate the concentration in grams per liter (g/L) of phosphoric acid in the rust remover.

HELP

H3PO4 + 3NaOH --> 3NaPO4 + 3H2O

n(NaOH) = c x v
n(NaOH) = 0.930mol/L x 21.3x10^-3L
n(NaOH) = 0.019809mol

n(H3PO4) = n(NaOH) x unknown/known
n(H3PO4) = 0.019809mol x 1/3
n(H3PO4) = 6.603x10^-3

dilute c(H3PO4) = n / v
dilute c(H3PO4) = 6.603mol / 20x10^-3L
dilute c(H3PO4) = 0.33015mol/L

original c(H3PO4) = dilute c(H3PO4) x dilution factor
original c(H3PO4) = 0.33015mol/L x 200/10
original c(H3PO4) = 6.603mol/L

original c(H3PO4) in g/L = original c(H3PO4) x M(H3PO4)
original c(H3PO4) in g/L = 6.603mol/L x 98g/mol
original c(H3PO4) in g/L = 647.094g/L

Therefore the original concentration of the Phosphoric acid in the rust remover is 647.094 grams per litre