Answer
The lower quantum number is $n=3$
Work Step by Step
An electron confined to an infinite potential well can exist in only certain discrete states. If the well is one-dimensional with length $L$, the energies associated with these quantum states are
$E_n=\Big(\frac{h^2}{8mL^2}\Big)n^2,\;\;\;\text{for}\;n=1,2,3....,$
where $m$ is the electron mass and $n$ is a quantum number.
The lowest energy corresponds to the ground state $(n=1)$ is,
$E_1=\frac{h^2}{8mL^2}$
Thus the energies associated with the quantum states of an electron confined to an infinite potential well can be written as
$E_n=n^2E_1$
Thus, the energy that corresponds energy level $n$ is: $E_n=n^2E_1$
and the energy that corresponds energy level $n+1$ is: $E_{n+1}=(n+1)^2E_1$
Therefore, the energy difference between the energy levels $n$ and $n+1$ becomes
$E_{n+1}-E_n=\{(n+1)^2-n^2\}E_1=(2n+1)E_1$
Now, the energy difference $E_{43}$ between the levels $n =4$ and $n=3$ is given by
$E_{43}=4^2E_1-3^2E_1=7E_1$
According to the given condition,
$E_{n+1}-E_n=E_{43}$
or, $(2n+1)E_1=7E_1$
or, $(2n+1)=7$
or, $n=3$
Therefore, the lower quantum number is $n=3$
