Answer
$r_3 = 4.77\times 10^{-10}~m$
Work Step by Step
We can find the orbital radius of the electron:
$r_n = \frac{n^2~h^2}{4~\pi^2~m_e~k~Z~e^2}$
$r_3 = \frac{(3)^2(6.626\times 10^{-34}~J~s)^2}{(4~\pi^2)(9.1\times 10^{-31}~kg)(9.0\times 10^9~N~m^2/C^2)(1)(1.6\times 10^{-19}~C)^2}$
$r_3 = 4.77\times 10^{-10}~m$
