Answer
Mass defect = 0.44947 amu
Nuclear binding energy = 8.723 MeV/nucleon
Work Step by Step
Mass defect=[Z(mass $^{1}_{1}H$)+(A-Z)(mass $^{1}_{0}n$)]-mass $^{48}_{22}Ti$
=22(1.00783 amu)+26(1.00866 amu)- 47.947947 amu
=0.44947 amu.
Binding energy=$0.44947\,amu\times\frac{931.5\,MeV}{1\,amu}=418.68\,MeV$
Nuclear binding energy per nucleon=
$\frac{Binding\,energy}{Number\,of\,nucleons (A)}=\frac{418.68\,MeV}{48\,nucleons}=8.723\,MeV/nucleon$
