= Solution
The original diagram places hard walls at $0$ and $3L$, and the central plateau on $L<x<2L$ is \b[exactly at the indicated ground-state <energy>]. This equality determines the middle of the requested sketch. In the <Time-independent Schrödinger equation>,
$$
\psi''=\frac{2m}{\hbar^2}(V-E)\psi,
$$
the central interval has $V=E$, hence $\psi''=0$. The <ground state> can be chosen real and positive inside the walls by the <nodeless theorem for a one-dimensional ground state>. Reflection symmetry makes it even about $x=3L/2$, so its linear central segment must be constant.
The hard-wall boundary conditions are $\psi(0)=\psi(3L)=0$. On both outer intervals $V=0<E$, so the positive <wavefunction> has negative curvature and is sinusoidal. <Wavefunction matching at a finite potential step> requires both $\psi$ and $\psi'$ to remain continuous at $L,2L$; the <derivative> therefore vanishes where each sinusoidal flank meets the flat middle. A normalized representative is
$$
\boxed{\psi_0(x)=\frac1{\sqrt{2L}}\begin{cases}
\sin\!\left(\dfrac{\pi x}{2L}\right),&0<x<L,\\
1,&L\le x\le2L,\\
\sin\!\left(\dfrac{\pi(3L-x)}{2L}\right),&2L<x<3L,\\
0,&x\le0\ \text{or}\ x\ge3L.
\end{cases}}
$$
Each side contributes $L/2$ and the plateau contributes $L$ to the squared norm before multiplying by $1/(2L)$. The corresponding consistency condition is $E=\pi^2\hbar^2/(8mL^2)$; a numerical <energy> is not required for the sketch.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-1-q1-flat-barrier-state.png]
{title=The positive ground-state wavefunction has sinusoidal flanks and a flat central segment at V equals E}
Thus the <ground state at the top of a central rectangular barrier> has \b[no internal nodes, continuous value and slope, and a flat middle], rather than an exponential dip through a classically forbidden region. Multiplication by an overall constant phase changes none of these physical features.
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