Let the original radius be $r$ and the original height be $h$.
The original volume of the cylinder, $\(V_1\)$, is given by the formula:

$$
\(V_1=\pi r^2 h\)
$$


The radius is increased by $\(20 \%\)$. So, the new radius, $\(r^{\prime}\)$, will be:

$$
\(r^{\prime}=r+0.20 r=1.20 r\)
$$


The height remains the same, so the new height is still $h$.
The new volume of the cylinder, $V_2$, will be:

$$
\(\begin{aligned}
& V_2=\pi\left(r^{\prime}\right)^2 h=\pi(1.20 r)^2 h \\
& V_2=\pi\left(1.44 r^2\right) h \\
& V_2=1.44 \pi r^2 h
\end{aligned}\)
$$


Now, we can express $\(V_2\)$ in terms of $\(V_1\)$ :
Since $\(V_1=\pi r^2 h\)$, we have $\(V_2=1.44 V_1\)$.
To find the percentage change in volume, we use the formula:

$$
\(\begin{aligned}
& \text { Percentage Change }=\frac{V_2-V_1}{V_1} \times 100 \% \\
& \text { Percentage Change }=\frac{1.44 V_1-V_1}{V_1} \times 100 \% \\
& \text { Percentage Change }=\frac{(1.44-1) V_1}{V_1} \times 100 \% \\
& \text { Percentage Change }=(0.44) \times 100 \% \\
& \text { Percentage Change }=44 \%
\end{aligned}\)
$$


Therefore, the percentage change in the volume of the cylinder is $\(44 \%\)$.
The correct answer is (C).