Binding energy of Li 6

Homework Statement

How do I find the binding energy for the nuclei of Lithium 6.
N=Z=3, atomic mass=6.015 u, binding energy per nucleon= 5.33 MeV
The equation I'm using is :

B(N,Z)=Z*m(0,1)*c^2+N*(mass of n)*c^2-m(N,Z)*c^2.

B(N,Z)=3*(1.00783u*1.66E-27kg)*(3E8 m/s)^2+3*(1.008664u*1.66E-27kg)*(3E8 m/s)^2-(6.015u**1.66E-27kg)*(3E8 m/s)^2

B(N,Z)= 4.84056E-12 J *6.24E18 eV= 30.2 MeV

but that's not the answer, I also tried dividing that answer by the binding energy per nucleon which is 5.332 MeV and I got 5.66 MeV that's not the answer. I'll appreciate it if someone explained to me what I'm doing wrong. Thanks.