In an experiment to determine the density of a cylindrical metal...

The absolute error in each measurement is equal to the least count of the respective measuring instrument.

For mass $M$:
$\Delta M = 0.01\text{ g}$
Percentage error in $M = \dfrac{0.01}{1.00} \times 100 = 1\%$

For length $L$:
Least count $= 1\text{ mm} = 0.1\text{ cm}$
$\Delta L = 0.1\text{ cm}$
Percentage error in $L = \dfrac{0.1}{10.0} \times 100 = 1\%$

For diameter $D$:
Least count of screw gauge $= \dfrac{\text{Pitch}}{\text{Number of circular divisions}} = \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}$
$\Delta D = 0.01\text{ mm}$
Percentage error in $D = \dfrac{0.01}{2.00} \times 100 = 0.5\%$

The density of a cylinder in terms of its diameter is given by
$\rho = \dfrac{M}{\pi \left(\dfrac{D}{2}\right)^2 L} = \dfrac{4M}{\pi D^2 L}$

The maximum percentage error in density is
$\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta M}{M} \times 100 + 2\left(\dfrac{\Delta D}{D} \times 100\right) + \dfrac{\Delta L}{L} \times 100$

Substitute the calculated percentage errors:
$\dfrac{\Delta \rho}{\rho} \times 100 = 1\% + 2(0.5\%) + 1\% = 1\% + 1\% + 1\% = 3\%$

Answer: $3\%$