Solution:

To find the dimensions of \(\frac{Kq^{2}}{Gm^{2}}\), we analyze the dimensions of each component.

1. Dimensions of \( K \):
- Given \( K = \frac{1}{4\pi \varepsilon_0} \), where \(\varepsilon_0\) is the permittivity of free space.
- \( K \) has dimensions of \(\left[ \text{Force} \cdot \text{Distance}^2 \cdot \text{Charge}^{-2} \right] = \left[ M^1 L^3 T^{-4} A^{-2} \right] \).

2. Dimensions of \( q^2 \):
- The charge \( q \) has dimensions \(\left[ A T \right]\), so \( q^2 \) has dimensions \(\left[ A^2 T^2 \right]\).

3. Dimensions of \( G \):
- \( G \) is the gravitational constant with dimensions \(\left[ M^{-1} L^3 T^{-2} \right]\).

4. Dimensions of \( m^2 \):
- The mass \( m \) has dimensions \(\left[ M \right]\), so \( m^2 \) has dimensions \(\left[ M^2 \right]\).

5. Combine Everything:
- Now, \(\frac{Kq^2}{Gm^2}\) has dimensions:
\[
\frac{\left[ M^1 L^3 T^{-4} A^{-2} \right] \cdot \left[ A^2 T^2 \right]}{\left[ M^{-1} L^3 T^{-2} \right] \cdot \left[ M^2 \right]}
\]

6. Simplify:
- This simplifies to \(\left[ M^0 L^0 T^0 A^0 \right] = \text{No Dimensions}\).

Conclusion:
The expression \(\frac{Kq^2}{Gm^2}\) is dimensionless.