Hope the image of the worked out problem attached helps. I'll walk through my brain jumble:

1) To solve for the KE of the ejected electron, use the equation on the right side of the diagram because this is the amount of KE of the electrons. This equation is E

Now, all you have to do is take your known quantities (velocity, mass of electron constant), then plug and chug. You should get

2) To solve for the energy needed to remove an e- from the metal, remember that this energy is the amount of energy the photon needs to meet or exceed to remove the e-, so wouldn't it make sense for this energy to be the threshold energy (work function)? Yes, it does. You are actually given the threshold energy, but it is in KJ. Convert the quantity into Joules by a unit conversion (1000 Joules in 1 KJ). You should get

3) Finally, we need to find the frequency of the incident light. You already know that

Using this equation, rearrange to solve for the energy of the photon (because we know the threshold and KE of photon already). Plug and chug to solve for the energy. Now that we know the energy, to find the frequency of the light, use the equation

Rearrange the equation to solve for frequency. Plug and chug, then you're done! Yippee! :)

https://ibb.co/ciCKXkThis is the image^ Idk how to post it. LMK if the URL fails me

**EDIT**

Divide the energy to remove the metal (threshold) by Avogadro's number because it is the energy per electron. Then, use that number instead of the mess I wrote to plug and chug to find the frequency.