To develop this problem we require the concepts related to wavelength and its expression to calculate it.
The wavelength is given by
[tex]\lambda =\frac{c}{f}[/tex]
Where,
[tex]c=3*10^8 m/s[/tex] light velocity
f = frequency.
Our values are given by,
[tex]f_1=4.38*10^{14}Hz[/tex]
[tex]f_2= 4.14*10^{20}Hz[/tex]
[tex]f_3 = 3.24*10^{12}Hz[/tex]
Then,
[tex]\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m[/tex] Visible
[tex]\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m[/tex] Gamma Ray
[tex]\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m[/tex] Infrared
*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.