You measured the mass of the water to be equal to 17.97 g, which means that you can use water's density at the temperature at which you did the measurements to figure out th volume that sample would occupy.
Density is defined as mass per unit of volume
#rho = n/V#
Assuming that you did the measurements at
The theoretical volume of the sample would be
#17.97color(red)(cancel(color(black)("g"))) * "1 mL"/(0.9982071color(red)(cancel(color(black)("g")))) = "18.002 mL"#
To get the percent error, use the formula
#color(blue)("% error" = (|"measured value" - "theoretical value"|)/"theoretical value" * 100)#
In your case, you have
#"% error" = (|(20.00 - 18.002)|color(red)(cancel(color(black)("mL"))))/(18.002color(red)(cancel(color(black)("mL")))) * 100#
#"% error" = 1.998/18.002 * 100 = color(green)("11.10 %")#
SIDE NOTE I used four significant figures for the measured volume to keep in line with the number of sig figs you gave for the mass of water.
If the measured volume is not 20.00, then you need to round the answer to the number of sig figs you have for the measured volume.