Answer
See work below.
Work Step by Step
Find the energy from the wavelength of the emitted photon.
$$E = hf=\frac{hc}{\lambda}$$
$$ =\frac{(6.63\times10^{-34}J \cdot s)(3.00 \times10^8 m/s)}{(1.15\times10^{-13}m)(1.60\times10^{-13}J/MeV)}$$
$$=10.8 MeV$$
This is a $\gamma$ ray from the nucleus, because electron transitions emit photons with much less energy than this (from a few to tens of eV).
