An electromagnetic wave has a frequency of and is traveling in a vacuum. the magnetic field is given by

(a) find the wavelength and the direction of propagation of this wave.
(b) find the electric field vector.
(c) determine the poynting vector, and use it to find the intensity of the wave.

c) true. With the magnetic field you can determine the Poynting vector and this is equal to the intensity of the wave

Explanation:

An electromagnetic wave carries energy that is given by the vector of poyntíng

     S = 1 / μ₀ E x B

The electric and magnetic fields are related

     E = c B

Substituting

     S = c /μ₀ B² / 2

The two comes from the average value

Let's examine the claims

a) false to determine the wavelength the frequency is needed

b) False. To find the electric field you need the magnetic field

c) true. With the magnetic field you can determine the Poynting vector and this is equal to the intensity of the wave

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m