The length of a metallic wire is increased by 20% and its area of cross-section is reduced by 4%. The percentage change in resistance of the metallic wire is ______.

The resistance of a metallic wire is given by:

$$R = \frac{\rho L}{A}$$

where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. We have the length increased by 20%, so the new length is $$1.2L$$, and the area reduced by 4%, so the new area is $$0.96A$$.

Now the new resistance is:

$$R' = \frac{\rho \cdot 1.2L}{0.96A} = \frac{1.2}{0.96} \cdot \frac{\rho L}{A} = 1.25R$$

The percentage change in resistance is:

$$\frac{R' - R}{R} \times 100 = (1.25 - 1) \times 100 = 25\%$$

So, the answer is $$25$$.