Four measurements of the mass of a block are recorded as 2.01 ~g ,...

First, calculate the mean of the mass measurements:
Sum $= 2.01 + 2.02 + 2.03 + 2.05 = 8.11 \mathrm{~g}$
Mean $= \dfrac{8.11}{4} = 2.0275 \mathrm{~g}$

Since the original measurements have two decimal places, the mean must be rounded to two decimal places. Rounding $2.0275 \mathrm{~g}$ gives $2.03 \mathrm{~g}$.

Next, calculate the absolute errors for each measurement using the rounded mean ($2.03 \mathrm{~g}$):
$|2.01 - 2.03| = 0.02 \mathrm{~g}$
$|2.02 - 2.03| = 0.01 \mathrm{~g}$
$|2.03 - 2.03| = 0.00 \mathrm{~g}$
$|2.05 - 2.03| = 0.02 \mathrm{~g}$

Sum of absolute errors $= 0.02 + 0.01 + 0.00 + 0.02 = 0.05 \mathrm{~g}$
Mean absolute error $= \dfrac{0.05}{4} = 0.0125 \mathrm{~g}$

The mean absolute error must also be rounded to two decimal places to match the precision of the mean. Rounding $0.0125 \mathrm{~g}$ to two decimal places gives $0.01 \mathrm{~g}$.

The final reported result is $2.03 \pm 0.01 \mathrm{~g}$.

Answer: $2.03 \pm 0.01 \mathrm{~g}$