For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible.
In option (b), $n_{1}=2, n_{2}=1$
$\therefore$ Energy of emitted photon is given as
$$
\begin{aligned}
E^{\prime} &=-13.6 Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\Rightarrow \quad h v^{\prime} &=-13.6 Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right) \\
v^{\prime} &=\frac{13.6 Z^{2}}{h} \times \frac{3}{4}
\end{aligned}
$$
In option (c), $n_{1}=2, n_{2}=6$, hence energy of emitted photon is given as
$$
\begin{aligned}
& E^{\prime \prime}=-13.6 Z^{2}\left(\frac{1}{6^{2}}-\frac{1}{2^{2}}\right) \\
\Rightarrow & h v^{\prime \prime}=13.6 Z^{2} \times \frac{8}{36} \\
\Rightarrow & \quad v^{\prime \prime}=\frac{13.6 Z^{2}}{h} \times \frac{8}{36}
\end{aligned}
$$
Hence, from Eqs. (i) and (ii), we get
$$
v^{\prime}>v^{\prime \prime}
$$