Answer
$1.0\;\rm nm$
Work Step by Step
For an electron in a one-dimensional box of length $ L $, the energy levels are quantized and given by
$$ E_n = \frac{n^2 h^2}{8mL^2} \tag1 $$
where $ n $ is the quantum number (integer), $ h $ is Planck's constant, $ m $ is the mass of the electron, $ L $ is the length of the box.
Solving for $L$;
$$ L^2 = \frac{n^2 h^2}{8mE_n} $$
$$ L = \frac{n h }{ \sqrt{8mE_n} }$$
We can see from the given graph that $n=4$, so
$$ L = \frac{4 h }{ \sqrt{8mE_4} }$$
Substitute the known:
$$ L = \frac{4 (6.63 \times 10^{-34}) }{ \sqrt{8 (9.11 \times 10^{-31}) (6\times 1.6\times 10^{-19})} }$$
$$L=\bf 1.002\times10^{-9}\;\rm m=\color{red}{\bf 1.0}\;\rm nm$$
