Step 1: Recall the photoelectric balance.
The incoming photon energy $h\nu$ splits into the work function $\Phi$ plus the electron's kinetic energy, $h\nu = \Phi + K_{\max}$. The work function itself is $\Phi = h\nu_0$ using the threshold frequency $\nu_0$.
Step 2: Find the work function.
With $\nu_0 = 1.15 \times 10^{15}$ Hz and $h = 6.60 \times 10^{-34}$ Js, \[ \Phi = (6.60 \times 10^{-34})(1.15 \times 10^{15}) = 7.59 \times 10^{-19} \text{ J} \]
Step 3: Convert the kinetic energy to joules.
$K_{\max} = 0.20$ eV, and $1$ eV $= 1.6 \times 10^{-19}$ J, so \[ K_{\max} = 0.20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-20} \text{ J} \]
Step 4: Add the two energies.
The numerator is $\Phi + K_{\max} = 7.59 \times 10^{-19} + 0.32 \times 10^{-19} = 7.91 \times 10^{-19}$ J.
Step 5: Divide by Planck's constant.
\[ \nu = \frac{7.91 \times 10^{-19}}{6.60 \times 10^{-34}} \approx 1.20 \times 10^{15} \text{ Hz} \]
Step 6: Pick the option.
The frequency is $1.20 \times 10^{15}$ Hz, which is option 2.
\[ \boxed{1.20 \times 10^{15} \text{ Hz}} \]