If the energy of an electron in the first Bohr orbit of a hydrogen...

The energy of an electron in the $n$-th Bohr orbit of a hydrogen-like species is given by $E_n = -E_0 \dfrac{Z^2}{n^2}$.

For a hydrogen atom ($Z=1$) in the first Bohr orbit ($n=1$), the energy is $E_1(\text{H}) = -E_0 \dfrac{1^2}{1^2} = -E$.

For a $\text{He}^{+}$ ion ($Z=2$) in the third Bohr orbit ($n=3$), the energy is $E_3(\text{He}^{+}) = -E_0 \dfrac{2^2}{3^2} = -E_0 \dfrac{4}{9}$.

Substituting $E_0 = E$, we get $E_3(\text{He}^{+}) = -\dfrac{4}{9}E$.

Answer: $-\dfrac{4}{9}E$