Answer
The electric potential at the center is $~~2.21~V$
Work Step by Step
We can write the general expression for the electric potential at a point due to a system of charges:
$V = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i}{r_i}$
Note that the four charges $+2q_1, +2q_1, -q_1,$ and $-3q_1$ are the same distance from the center. Therefore the electric potential due to these four charge will cancel out since the sum of the charges is zero.
We can find the electric potential at the center:
$V = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i}{r_i}$
$V = \frac{1}{4\pi~\epsilon_0}~\frac{8q_2}{a/2}$
$V = \frac{1}{\pi~\epsilon_0}~\frac{4q_2}{a}$
$V = \frac{1}{(\pi)(8.854\times 10^{-12}~C/V~m)}~\frac{(4)(6.00\times 10^{-12}~C)}{0.39~m}$
$V = 2.21~V$
The electric potential at the center is $~~2.21~V$
