Answer
$V = 4.49\times 10^4~V$
Work Step by Step
The electric potential inside a spherical shell due to the charge on the outside of the shell is equal to the electric potential on the surface of the shell. That is $V = \frac{1}{4\pi~\epsilon_0}~\frac{Q}{R}$
The electric potential outside a spherical shell due to the charge on the outside of the shell is $~~V = \frac{1}{4\pi~\epsilon_0}~\frac{Q}{r}~~$ where $r$ is the distance from the center.
We can find the electric potential at $r = 0$:
$V = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1}{R_1}+\frac{q_2}{R_2})$
$V = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~(\frac{2.00\times 10^{-6}~C}{0.500~m}+\frac{1.00\times 10^{-6}~C}{1.00~m})$
$V = 4.49\times 10^4~V$
