Answer
$\lambda = 1.06\times 10^{-12}~m$
Work Step by Step
We can find $\gamma$ for the electron and positron:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{(0.9~c)^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-0.81}}$
$\gamma = 2.294$
We can find the total energy for the electron:
$E = \gamma~mc^2$
$E = (2.294)(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$
$E = 1.88\times 10^{-13}~J$
Note that the positron also has the same total energy.
We can find the wavelength of each photon with this energy:
$E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{1.88\times 10^{-13}~J}$
$\lambda = 1.06\times 10^{-12}~m$
