= Solution
The <electric potential difference> is the negative integral of the field:
$$
V_b-V_{\rm ext}=-\int_{-w/2}^{w/2}E_x(x)\,dx
=\frac{e\alpha}{2\varepsilon_s}\int_{-w/2}^{w/2}\left(\frac{w^2}{4}-x^2\right)dx
=\frac{e\alpha w^3}{12\varepsilon_s}.
$$
Consequently
$$
\boxed{w(V_{\rm ext})=\left[\frac{12\epsilon_0\epsilon(V_b-V_{\rm ext})}{e\alpha}\right]^{1/3},\qquad
w(0)=\left(\frac{12\epsilon_0\epsilon V_b}{e\alpha}\right)^{1/3}.}
$$
The <linearly graded junction depletion width> has this cube-root dependence because of the linearly growing space charge: the charge on one side scales as $w^2$, the field as $w^2$, and its potential integral as $w^3$. The formula assumes $V_b-V_{\rm ext}>0$ and that the <depletion approximation> remains applicable.
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