Answer
$0.999c$
Work Step by Step
We know that
$v_e^2=\frac{(1836)(v_p)}{\sqrt{1+((1836)^2-1)(\frac{v_p^2}{c^2})}}$
We plug in the known values to obtain:
$v_e^2=\frac{(1836)(0.0100c)}{\sqrt{1+((1836)^2-1)(\frac{0.0100c}{c})^2}}$
This simplifies to:
$v_e=0.999c$
