hc     6.63 * 10^(-34) J sec * 3 * 10^17 nm/s
(a) Energy = ------ = --------------------------------------
             Lambda                350 nm

           = 5.68 * 10^-19 Joule

(b) The wavelength of a photon with 2.1 * 10^(-18) Joule is given by

               hc     6.63 * 10^(-34) J sec * 3 * 10^17 nm/s
    Lambda = ------ = -------------------------------------- = 94.7 nm
             Energy             2.1 * 10^(-18) J

From the equation, you can see that the larger the wavelength gets, the
smaller the energy gets.  Thus, the maximum wavelength of a hydrogen
ionizing photon here is 94.7 nm.  Also from the equation, the smaller
the wavelength, the larger the energy; since you only need a photon with
more energy than 2.1 * 10^(-18) J, the minimum wavelength of a hydrogen
ionizing photon is zero.  You can also reason that the minimum wavelength
would the minimum wavelength allowed by nuclear processes; this is not
necessary, but just for reference, a photon with a wavelength shorter
than about 0.001 nm will often turn into matter and antimatter spontaneously.

(c) Using Wien's Law, we know that Lambda * Temperature = 2.9 * 10^6 nm K.
    So for Lambda = 94.7 nm,  T = 2.9 * 10^6 / 94.7 = 30600 K.  The sun's
    surface temperature is about 5800 K, so such an object would have a
    surface temperature over five times that of the Sun.