Step-by-Step Solution

1. Identify the frequencies present in the electric field equation: $E = 60[\sin(3\times10^{14}t)+\sin(12\times10^{14}t)]$. The angular frequencies are $\omega_1 = 3\times10^{14}$ rad/s and $\omega_2 = 12\times10^{14}$ rad/s. The corresponding frequencies are $f_1 = \omega_1 / 2\pi$ and $f_2 = \omega_2 / 2\pi$.

2. Calculate the energy of the photons corresponding to these frequencies using $E = hf = \hbar\omega$, where $h$ is Planck's constant ($h \approx 6.626 \times 10^{-34} Js$) and $\hbar = h/2\pi$. Also, $h = 4.135 \times 10^{-15} eV \cdot s$.

   For $\omega_1 = 3\times10^{14}$ rad/s: $E_1 = \hbar\omega_1 = (4.135 \times 10^{-15} eV \cdot s)(3\times10^{14}) = 1.2405 eV$.

   For $\omega_2 = 12\times10^{14}$ rad/s: $E_2 = \hbar\omega_2 = (4.135 \times 10^{-15} eV \cdot s)(12\times10^{14}) = 4.962 eV$.

3. Apply the photoelectric effect equation: $KE_{max} = E - W$, where $KE_{max}$ is the maximum kinetic energy of the ejected electrons, $E$ is the energy of the incident photon, and $W$ is the work function of the material.

4. Since we want the *maximum* kinetic energy, we should use the higher energy photon ($E_2 = 4.962 eV$). The work function is given as $W = 2.8 eV$.

   $KE_{max} = E_2 - W = 4.962 eV - 2.8 eV = 2.162 eV$.

5. Round the result to two decimal places: $KE_{max} \approx 2.16 eV$.