1. Identify the frequencies present in the light wave

   The magnetic field is $B = B_0\left[\sin\left((3.14\times 10^7)ct\right)+\sin\left((6.28\times 10^7)ct\right)\right].$

   Comparing with the standard form $\sin(\omega t),$ the angular frequencies are $\omega_1=(3.14\times 10^7)c,$ $\omega_2=(6.28\times 10^7)c.$

   Given $c=3\times 10^8\ \text{m s}^{-1},$ we get $\omega_1=3.14\times 10^7\times 3\times 10^8=9.42\times 10^{15}\ \text{rad s}^{-1},$ $\omega_2=6.28\times 10^7\times 3\times 10^8=1.884\times 10^{16}\ \text{rad s}^{-1}.$

2. Find the corresponding frequencies

   Using $\nu=\frac{\omega}{2\pi},$

   for the first component, $\nu_1=\frac{9.42\times 10^{15}}{2\times 3.14}=1.5\times 10^{15}\ \text{Hz}.$

   for the second component, $\nu_2=\frac{1.884\times 10^{16}}{2\times 3.14}=3.0\times 10^{15}\ \text{Hz}.$

3. Photoelectric effect depends on the highest frequency

   Since the light contains two frequencies, the maximum kinetic energy of emitted photoelectrons will be due to the photon with larger frequency: $\nu_{\max}=3.0\times 10^{15}\ \text{Hz}.$

4. Calculate the maximum photon energy

   $E=h\nu=(6.6\times 10^{-34})(3.0\times 10^{15})=19.8\times 10^{-19}\ \text{J}.$

   Convert to eV using $1\ \text{eV}=1.6\times 10^{-19}\ \text{J},$ so $E=\frac{19.8\times 10^{-19}}{1.6\times 10^{-19}}=12.375\ \text{eV}.$

5. Apply Einstein’s photoelectric equation

   $K_{\max}=h\nu-\phi$ where the work function of silver is $\phi=4.7\ \text{eV}.$

   Therefore, $K_{\max}=12.375-4.7=7.675\ \text{eV}.$

   This is closest to $\boxed{7.72\ \text{eV}}.$

6. Check options

   • A: $6.82\ \text{eV}$ ❌
   • B: $12.5\ \text{eV}$ ❌ (this is approximately photon energy, not kinetic energy)
   • C: $8.52\ \text{eV}$ ❌
   • D: $7.72\ \text{eV}$ ✅

Hence the correct option is D.