equivalent load resistance
The output current value cited is the average current of a sine wave, which mathemagically is two over pi times the peak value. This can be proven by integrating I * sin(x) from zero to pi, multiplying by 2 to mimic the rectification (both peaks in the same direction) and dividing by 2 * pi since we're dealing with a whole AC cycle, and we get our 2 over pi factor.

EDIT: A derivation (calculus = rusty, apologizes for glaring errors)

\$ I_o = 2\dfrac{\int\limits_{0}^{\pi}{I_{ac}}^{pk}\cdot \textrm{sin}(x) \textrm{d}x}{2\pi} \$

\$ I_o = 2\dfrac{{I_{ac}}^{pk}\cdot \Big(-\textrm{cos}(\pi) - \big(-\textrm{cos}(0)\big)\Big)}{2\pi}\$

\$ I_o = 2\dfrac{{I_{ac}}^{pk}\cdot \Big(-(-1) + (1)\Big)}{2\pi}\$

\$ I_o = 2\dfrac{2{I_{ac}}^{pk}}{2\pi}\$

\$ I_o = \dfrac{2{I_{ac}}^{pk}}{\pi}\$