Error in the measurement of radius of the sphere is $2 \%$. The error in the calculated value of its volume is

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.

Since $V$ is proportional to $r^3$, the percentage error in volume relates to the percentage error in radius by the power rule:
$\frac{\Delta V}{V} \times 100\% = 3 \left( \frac{\Delta r}{r} \times 100\% \right)$

Given that the radius has a $2\%$ error, the volume error becomes:
$\frac{\Delta V}{V} \times 100\% = 3 \times 2\% = 6\%$

Answer: $\boxed{\text{C}}$