From the given tabular data,
Volume of water (V) = 50.0 ml
Mass of water (m) = 49.7 g
So, the density of water (ρ) = Mass / Volume
or, = 49.7 g /50.0 ml = 0.994 g / ml
<u>Actual value of the density of water (ρ') = 0.998 g /ml</u>
So, the percentage error in the density of water
= [( actual value - observed value) / actual value ] ×100
or, = [ (0.998 g /ml - 0.994 g / ml ) / 0.998 g /ml ] ×100
or, = 0.4 %
Hence, the required % error in the density of water will 0.4 %.