Answer
${\bf 200 }\; \rm {nm}$
Work Step by Step
We know that
$$ E_n = \left(n - \frac{1}{2}\right)\hbar \omega_e\tag {where $n=1,2,3,...$}$$
So, the energy of the photon absorbed in the $1\rightarrow2$ quantum jump is
$$E_2 - E_1 = \left(2 - \frac{1}{2}\right)\hbar \omega_e - \left(1 - \frac{1}{2}\right)\hbar \omega_e $$
$$ E_2 - E_1 =\hbar \omega_e = \frac{hc}{\lambda_{2\rightarrow 1}} \tag 1$$
And hence,
$$E_3 - E_1 = \left(3 - \frac{1}{2}\right)\hbar \omega_e - \left(1 - \frac{1}{2}\right)\hbar \omega_e $$
$$E_3 - E_1= 2\hbar \omega_e = \frac{hc}{\lambda_{3\rightarrow 1}}\tag 2 $$
From (1) and (2), we can see that
$$ \lambda_{31} = \frac{\lambda_{21}}{2} = \frac{400 \; \rm {nm}}{2} $$
$$ \lambda_{31} =\color{red}{\bf 200 }\; \rm {nm}$$
