The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by:

$E = K_{\max} + W$,

where

$E$ is the energy of the incident light,

$K_{\max}$ is the maximum kinetic energy of the photoelectrons, and

$W$ is the work function of the metal.

The energy of the incident light can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. However, given that $hc = 1240$ eV⋅nm, we can simplify this to $E = \frac{1240}{\lambda}$.

Substituting the given values into this equation, we have:

$E = \frac{1240}{400} = 3.1$ eV.

We can then substitute these values into the photoelectric effect equation:

$3.1 \text{ eV} = 1.68 \text{ eV} + W$,

which simplifies to:

$W = 3.1 \text{ eV} - 1.68 \text{ eV} = 1.42$ eV.

Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.

Thus, Option A: $1.41$ eV is the closest to the correct answer.