Answer
$\lambda_i = 121.8~pm$
Work Step by Step
We can find the Compton shift in wavelength:
$\Delta \lambda = \frac{h}{mc}~(1-cos~\theta)$
$\Delta \lambda = \frac{6.626\times 10^{-34}~J~s}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1-cos~100.0^{\circ})$
$\Delta \lambda = (2.427~pm)~(1.17365)$
$\Delta \lambda = 2.848~pm$
We can find the wavelength of the incident photons:
$\Delta \lambda = \lambda_f-\lambda_i$
$\lambda_i = \lambda_f-\Delta \lambda$
$\lambda_i = (124.65~pm)- (2.848~pm)$
$\lambda_i = 121.8~pm$
