The correct option is 1.6 × 10¹⁵ Hz.

To find the threshold frequency for the photoelectric effect, we use the relationship between the work function of a metal and its threshold frequency:
$\Phi =h{\nu }_0$
where:
• $\Phi$ is the work function of the metal,
• $h$ is Planck's constant, and
• ${\nu }_0$ is the threshold frequency.

Step 1: Convert the work function from electron-volts (eV) to Joules (J)
The given work function is:
$\Phi =6.63\text{ eV}$
Since $1\text{ eV}=1.6\times {10}^{-19}\text{ J}$, we have:
$\Phi =6.63\times 1.6\times {\mathrm{10}}^{-19}\text{ J}$

Step 2: Use Planck's constant to calculate the threshold frequency
Planck's constant, $h$, is approximately:
$h=6.63\times {\mathrm{10}}^{-34}\text{ J·s}$
Rearranging the formula to solve for the threshold frequency ${\nu }_0$:
${\nu }_0=\frac{\Phi }{h}$

Substituting the values into the equation:
${\nu }_0=\frac{6.63\times 1.6\times {\mathrm{10}}^{-19}}{6.63\times {\mathrm{10}}^{-34}}$

We can simplify the fraction by canceling out the $6.63$ term in both the numerator and the denominator:
${\nu }_0=1.6\times {\mathrm{10}}^{-19-(-34)}$
${\nu }_0=1.6\times {\mathrm{10}}^{15}\text{ Hz}$

Thus, the threshold frequency for the photoelectric effect is $1.6\times {10}^{15}\text{ Hz}$.