Answer
At the center the potential is $72.0 \,\text{V}$
Work Step by Step
The potential inversely proportional to the distance
$$ V \propto \dfrac{1}{r}$$
So, for inside the sphere, the potential is $V_1$ and the distance is $R$ and outside the sphere, the potential is $V_2$ and the distance is $r$
\begin{gather*}
\dfrac{V_{1}}{V_{2}} = \dfrac{r}{R} \\
V_1 = \left(\dfrac{r}{R}\right) V_2
\end{gather*}
Subsitute the values for $r, R$ and $V_2$
\begin{align*}
V_1 &= \left(\dfrac{r}{R}\right) V_2 \\
&= \left(\dfrac{1.2 \,\text{m}}{0.400 \,\text{m}}\right) (24.0 \,\text{V}) = \boxed{72.0 \,\text{V}}
\end{align*}
