Answer
$E = 1.09 \times 10^{-12} J = 6.8 MeV$
Work Step by Step
The equation to find radius of an atom is $r = r_0 A^{1/3} $
$r_{\alpha} = (1.2 \times 10^{-15} m) (4)^{1/3} = 1.90 \times 10^{-15} m $
$r_{Al} = (1.2 \times 10^{-15} m) (27)^{1/3} = 3.60 \times 10^{-15} m $
Distance between centers of nuclei when the atom touches each other is
$r = r_{\alpha} + r_{Al} = 1.90 \times 10^{-15} m + 3.60 \times 10^{-15} m = 5.50 \times 10^{-15}m $
Now we can find the energy required,
$E = \frac{q_1q_2}{4 \pi \epsilon_or} = k \frac{q_1q_2}{r} $
$E = (8.99 \times 10^9 N.m^3/C^2) \frac{(2 \times 1.602 \times 10^{-19} J) (13 \times 1.602 \times 10^{-19} J)}{5.50 \times 10^{-15}m }$
$E = 1.09 \times 10^{-12} J = 6.8 MeV$
