If the radius of the innermost Bohr orbit of a hydrogen atom is 53...

The radius of the $n^{\text{th}}$ Bohr orbit for a hydrogen-like ion is given by:
$r_n = r_1 \left( \dfrac{n^2}{Z} \right)$
where $r_1$ is the radius of the innermost orbit of the hydrogen atom, $n$ is the principal quantum number, and $Z$ is the atomic number.

For a singly ionised helium atom ($\text{He}^{+}$), $Z = 2$.
For the second allowed orbit, $n = 2$.

Substituting the values:
$r_2 = 53 \times \dfrac{2^2}{2}$
$r_2 = 53 \times 2 = 106\text{ pm}$

Answer: $106\text{ pm}$