Answer
The speed of the recoiling hydrogen atom is $v=4.01 ms^{-1}$
Work Step by Step
The energies of different levels in the hydrogen atom in the Bohr model is given by the equation (given in the book):
$$E_{n}=\frac{-13.61 eV}{n^{2}}$$
where $n$ refers to the energy level quantum number. For the ground state $n=1$.
Hence the photon energy released in a transition from $n=4$ to $n=1$ is
$$\Delta E=E_{4}-E_{1}=(-13.61)(\frac{1}{4^{2}}-\frac{1}{1^{2}})=12.72eV$$
in Joules that is $$\Delta E=2.03*10^{-18}J$$
This will be the energy of the photon emitted due to the transition.
The atom will recoil from the photon due to momentum conservation. Since the entire system was at rest before the photon emission, when the photon is emitted in one direction, the atom must recoil in the opposite direction to conserve the system momentum, analogous to a gun firing a bullet and recoiling.
Hence to conserve the total momentum, the momentum of the emitted photon has to be equal to the momentum of the recoiling atom. The momentum of the photon $p_{p}$ is given by
$$E=p_{p}c$$
where $E$ is the energy of the photon, and $c$ is the speed of light.
Hence $$p_{p}=6.7*10^{-27} kg\ ms^{-1}$$
Thus the momentum of recoiling atom $p_{A}$ is
$$p_{A}=6.7*10^{-27} kg\ ms^{-1}=mv$$
where $m$ is the mass of the atom and $v$ is its velocity.
The mass of a hydrogen atom is $$m=1.6735575*10^{-27} kg$$
(You can also just add the mass of a proton and electron to get the mass of the hydrogen atom).
Thus, $$6.7*10^{-27} kg\ ms^{-1}=v*1.67*10^{-27} kg$$
Hence the speed of the recoiling hydrogen atom is
$$v=4.01 ms^{-1}$$
