Answer
To accelerate the alpha particle, the potential should be $4.16 \times 10^{6} \mathrm{V}$
Work Step by Step
The potenatial difference that accelrates the alpha particle could be found by knowing the distance. The distance here will be $r = d+R$ where $R$ is the radius of the nucleus and $d$ is the distance outside the nucleus, $r$ is the distance between the center of the nucleus and alpha particle
\begin{align}
V & = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{(r)} \\
& =\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{(R+d)} \\
&=\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{79 \times 1.6 \times 10^{-19} \mathrm{C}}{7.3 \times 10^{-15} \mathrm{m}+2.0 \times 10^{-14} \mathrm{m}}\right) \\
&=\boxed{ 4.16 \times 10^{6} \mathrm{V}}
\end{align}
