Answer
$0.151nJ$
Work Step by Step
We can find the required energy of each of the two gamma rays as follows:
$2E_{gamma}=m_{proton}c^2+m_{anti-proton}c^2$
We plug in the known values to obtain:
$2E_{gamma}=(1.673\times 10^{-27}Kg)(3\times 10^8m/s)^2+(1.673\times 10^{-27}Kg)(3\times 10^8m/s^2)$
$2E_{gamma}=15.075\times 10^{-11}=0.151nJ$
