1. Identify the frequencies present in the light wave

The electric field is $E = 200\,[\sin(6\times 10^{15} t) + \sin(9\times 10^{15} t)]\ \text{V m}^{-1}$

So the light contains two angular frequencies: $\omega_1 = 6\times 10^{15}\ \text{rad s}^{-1}, \qquad \omega_2 = 9\times 10^{15}\ \text{rad s}^{-1}$

For photoelectric emission, the maximum kinetic energy comes from the photon with the higher frequency.

2. Convert angular frequency to photon energy

Using $E_{\gamma} = h\nu = h\frac{\omega}{2\pi}$

Given: $h = 4.14\times 10^{-15}\ \text{eV s}$

For the higher angular frequency $\omega_2 = 9\times 10^{15}\ \text{rad s}^{-1}$, $\nu_2 = \frac{\omega_2}{2\pi} = \frac{9\times 10^{15}}{2\pi}$

Hence photon energy: $E_2 = h\nu_2 = 4.14\times 10^{-15}\times \frac{9\times 10^{15}}{2\pi}$

$E_2 = \frac{4.14\times 9}{2\pi} \text{ eV}$

$E_2 = \frac{37.26}{6.283} \approx 5.93\ \text{eV}$

3. Apply Einstein's photoelectric equation

$K_{\max} = E_{\gamma} - \phi$

Given work function: $\phi = 2.50\ \text{eV}$

So, $K_{\max} = 5.93 - 2.50 = 3.43\ \text{eV}$

4. Match with the given options

$K_{\max} \approx 3.42\ \text{eV}$

So the correct option is: D: 3.42 eV

5. Check lower-frequency component (for completeness)

For $\omega_1 = 6\times 10^{15}$, $E_1 = 4.14\times 10^{-15}\times \frac{6\times 10^{15}}{2\pi} = \frac{24.84}{6.283} \approx 3.95\ \text{eV}$

Then, $K_{\max,1} = 3.95 - 2.50 = 1.45\ \text{eV}$

This is smaller, so the overall maximum remains from the higher frequency component.

Derived answer: Option D