A hypothetical hydrogen-like atom consists of a proton and a...

The energy of an electron in a Bohr orbit is directly proportional to the mass of the orbiting particle: $E \propto m$.

Let the mass of the unknown particle be $m_x$. The ratio of its mass to the electron's mass is:
$\dfrac{m_x}{m_e} = \dfrac{E_x}{E_e} = \dfrac{-2720 \text{ eV}}{-13.6 \text{ eV}} = 200$

Thus, $m_x = 200 m_e$.

The radius of a Bohr orbit is given by $r_n \propto \dfrac{n^2}{m}$.

For the first orbit ($n=1$) of the normal hydrogen atom:
$r_1 = 0.53 \text{ \AA}$

For the second orbit ($n=2$) of the hypothetical atom, the radius is:
$r_2' = r_1 \times \left(\dfrac{n^2}{1^2}\right) \times \left(\dfrac{m_e}{m_x}\right)$

$r_2' = 0.53 \times 4 \times \left(\dfrac{1}{200}\right) \text{ \AA}$

$r_2' = \dfrac{2.12}{200} \text{ \AA} = 0.0106 \text{ \AA}$

Converting to meters ($1 \text{ \AA} = 10^{-10} \text{ m}$):
$r_2' = 0.0106 \times 10^{-10} \text{ m} = 1.06 \times 10^{-12} \text{ m}$.

Answer: $1.06 \times 10^{-12} \text{ m}$