Answer
See the answer below.
Work Step by Step
From the text:
$IE=h\nu-KE$
$h.c/\lambda=3.40×10^{-18}\ J$
$IE=2.98×10^{-18}\ J$
$2.98×10^{-18}\ J\div 1.602×10^{-19}\ J/eV=18.6\ eV$
$2.98×10^{-18}\ J×6.022×10^{23}\ 1/mol=$
$1794.0\ kJ/mol$