Current drawn from source by buck reg
This is correct, or close:

Efficiency is \$\eta = \frac{P_{OUT}}{P_{IN}} \$ so \$P_{IN} = \frac{P_{OUT}}{\eta} \$ = 0.235 for \$\eta = 0.85\$  

The distinction I'm making is that \$P_{IN}\$ is not 1.15 \$\cdot P_{OUT}\$- it makes a much bigger difference if the efficiency is &lt;&lt; 1. 

Also note that if the buck regulator is capable of much higher power (say 2-3A) it may not be 85% efficient at 60mA out- maybe only 50%. 

In that case, the input current would be 40mA (you'd get 30mA if you used 1.5 * Pin).