Answer
$V=-2.3V$
Work Step by Step
We know that:
$V=\frac{1}{4\pi \epsilon_{\circ}}(\frac{Q_1}{R}-\frac{6Q_1}{R})$
$V=-\frac{1}{4\pi \epsilon_{\circ}}\frac{5Q_1}{R}$
We plug in the known values to obtain:
$V=-8.99\times 10^9(\frac{5(4.20\times 10^{-12})}{8.2\times 10^{-2}})=-2.3V$
