Answer
$1.0\times10^4 m$.
Work Step by Step
The angular half-width of the flashlight beam's central maximum, shining through an aperture D, is given in the middle of page 729, the same as equation 25–7.
$$\theta=\frac{1.22\lambda}{D}$$
$$=\frac{1.22(550\times 10^{-9}m) }{(0.050m)}=1.3\times10^{-5}\;rad$$
By the definition of the radian, the radius of the circular spot when the beam strikes the moon is $\theta L$, where L is the distance from the earth to the moon. The diameter d of the beam when it reaches the moon is two times the radius of the circular spot.
$$d=2(\theta L)=2(1.3\times10^{-5}\;rad)(3.84\times10^8 m)= 1.0\times10^4 m $$