= Solution
Treat the $p^+$ strips as much more strongly <Semiconductor doping>[doped] than the $n$ bulk, so the <Depletion region>[depletion layer] lies predominantly in the latter. Fully ionized <Donor impurities>[donors] leave uniform <space charge> $\rho=eN_D$. In one dimension,
$$
\frac{dE}{dx}=\frac{eN_D}{\epsilon_{\mathrm{Si}}},\qquad \epsilon_{\mathrm{Si}}=\epsilon_r\epsilon_0.
$$
At the depletion boundary $x=d$, the minimum-bias solution has $E(d)=0$. Thus $E(x)=eN_D(x-d)/\epsilon_{\mathrm{Si}}$, and the magnitude of the potential drop is
$$
\boxed{V_{\mathrm{dep}}=\left|\int_0^dE(x)dx\right|=\frac{eN_Dd^2}{2\epsilon_r\epsilon_0}.}
$$
For $N_D=10^{18}\,\mathrm{m^{-3}}$, $d=300\,\mu\mathrm m$, and $\epsilon_r=11.6$, this is
$$
\boxed{V_{\mathrm{dep}}\simeq70.2\,\mathrm V.}
$$
The applied <reverse bias> is approximately this value; more precisely the built-in <Built-in potential>[junction potential] contributes to the total <One-sided depletion voltage>[depletion voltage] and is subtracted from the required external bias. It is small on the scale of seventy volts and is not specified. Here the stated free-carrier density is used as the effective <Donor impurity>[donor density] of the neutral bulk before depletion.
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