Answer
$E = 84~MeV$
Work Step by Step
We can find the nuclear diameter of $^{238}U$:
$d = 2r_0A^{1/3}$
$d = (2)(1.2~fm)(238)^{1/3}$
$d = 14.87~fm$
We can find the energy of an alpha particle with this de Broglie wavelength:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{14.87 \times 10^{-15}~m}$
$E = 1.34\times 10^{-11}~J$
$E = (1.34\times 10^{-11}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 8.4\times 10^7~eV$
$E = 84~MeV$
