The kinetic energy (\( KE \)) of a monoatomic gaseous molecule can be calculated using the formula:

\[ KE = \frac{3}{2} kT \]

where:
\( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \, \text{J/K} \)),
\( T \) is the temperature in Kelvin.

Given that the kinetic energy is \( 0.414 \, \text{eV} \), we need to convert it to joules since the Boltzmann constant is in joules:

\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]

So, \( 0.414 \, \text{eV} = 0.414 \times 1.602 \times 10^{-19} \, \text{J} \)

\[ = 6.627 \times 10^{-20} \, \text{J} \]

Now, we can set up the equation:

\[ 6.627 \times 10^{-20} \, \text{J} = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times T \]

Solving for \( T \):

\[ T = \frac{6.627 \times 10^{-20}}{\frac{3}{2} \times 1.38 \times 10^{-23}} \]

\[ T \approx 3198 \, \text{K} \]

Therefore, the temperature at which the kinetic energy of the monoatomic gaseous molecule is equal to \(0.414 \, \text{eV}\) is approximately \(3198 \, \text{K}\).

So, the correct option is (a) \(3198 \, \text{K}\).