Answer
$V=18 \mathrm{~V}$
Work Step by Step
Now $r=\sqrt{x^2+d^2}$ where $d=0.15 \mathrm{~m}$, so that
$
V=\frac{1}{4 \pi \varepsilon_0} \int_0^{0.20} \frac{b x d x}{\sqrt{x^2+d^2}}=\left.\frac{b}{4 \pi \varepsilon_0}\left(\sqrt{x^2+d^2}\right)\right|_0 ^{0.20}\\=18 \mathrm{~V}
$
