Answer
$3.43\times 10^{-10}m$.
Work Step by Step
Figure 40.15b in the textbook shows values for the bound state energy of a square well in the situation $U_o=6E_{1-IDW}$.
We know $ E_{1-IDW}=\frac{ h^2}{8mL^2}$.
From the figure we see that $E_1=0.625E_{1-IDW}=0.625\frac{ h^2}{8mL^2}$.
Solve for the width L, using $E_1=2.00eV=3.20\times10^{-19}J$.
$$3.20\times10^{-19}J =0.625\frac{ h^2}{8mL^2}$$
$$L=3.43\times 10^{-10}m$$
