Answer
$V=1.86 \times 10^{-2} \mathrm{~V}$
Work Step by Step
Consider an infinitesimal segment of the rod, located between $x$ and $x+d x$. It has length $d x$ and contains charge $d q=\lambda d x=c x d x$. Its distance from $P_1$ is $d+x$ and the potential it creates at $P_1$ is
$
d V=\frac{1}{4 \pi \varepsilon_0} \frac{d q}{d+x}=\frac{1}{4 \pi \varepsilon_0} \frac{c x d x}{d+x} .
$
To find the total potential at $P_1$, we integrate over the length of the rod and obtain
$
\begin{aligned}
V & =\frac{c}{4 \pi \varepsilon_0} \int_0^L \frac{x d x}{d+x}=\left.\frac{c}{4 \pi \varepsilon_0}[x-d \ln (x+d)]\right|_0 ^L=\frac{c}{4 \pi \varepsilon_0}\left[L-d \ln \left(1+\frac{L}{d}\right)\right] \\
& =\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right)\left(28.9 \times 10^{-12} \mathrm{C} / \mathrm{m}^2\right)\left[0.120 \mathrm{~m}-(0.030 \mathrm{~m}) \ln \left(1+\frac{0.120 \mathrm{~m}}{0.030 \mathrm{~m}}\right)\right] \\
& =1.86 \times 10^{-2} \mathrm{~V}
\end{aligned}
$
