Answer
$4.6\times 10^4m/s$
Work Step by Step
We can find the required escape speed of electron as follows:
$v_e=\sqrt{\frac{2Kqq_{\circ}}{mr}}$
We plug in the known values to obtain:
$v_e=\sqrt{\frac{2(9\times 10^9N.m^2/C^2)(1.8\times 10^{-15}C)(1.6\times 10^{-19}C)}{(9.1\times 10^{-31}Kg)(2.7\times 10^{-3}m)}}$
$v_e=4.6\times 10^4m/s$
