An electron in a hydrogen atom is in a state with a total energy of...

The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by:
$E_n = -\dfrac{13.6}{n^2}\text{ eV}$

Given $E_n = -1.51\text{ eV}$:
$-1.51 = -\dfrac{13.6}{n^2}$
$n^2 = \dfrac{13.6}{1.51} = 9$
$n = 3$

The radius of the $n^{\text{th}}$ orbit is given by:
$r_n = a_0 n^2$
where $a_0$ is the ground state radius.

Substituting $n = 3$ and $a_0 = 0.53\text{ \AA}$:
$r_3 = 0.53 \times 3^2$
$r_3 = 0.53 \times 9 = 4.77\text{ \AA}$

Answer: $4.77\text{ \AA}$