how to explain (b)?

The question has already told you to use "mass defect", hence,

Mass defect, Md = final mass - initial mass
i.e. Md = M(Sr) + M(Ba) - M(Pu)
where M(Sr), M(Ba) and M(Pu) are the masses of strontium, barium and plutonium respectively.

Multiply both sides by the square of the speed of light (c^2),
(Md)c^2 = [M(Sr)]c^2 + [M(Ba)]c^2 - [M(Pu)]c^2

Using then mass-energy equation, E = mc^2
E = E(Sr) + E(Ba) - E(Pu)
where E = (Md)c^2 is the energy released due to mass defect Md.
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I give below the more rigorous solution:
E(Pu) = (145n + 94p) - M(Pu)
E(Sr) = (52n + 38p) - M(Sr)
E(Ba) = (92n + 56p) - M(Ba)
where n and p denote "neutron" and "proton" respectively.
Energy released E = [M(Pu) + n] - [M(Sr) + M(Ba) + 2n]
i.e. E = [145n+94p - E(Pu)] - [52n+38p -E(Sr)] - [92n+56p - E(Ba)] - n
E = -E(pu) + E(Sr) + E(Ba)
or E = E(Sr) + E(Ba) - E(Pu)

You can see that from the first 3 equations, the binding energy is related to the mass of the nucleus. Since energy released is related to mass defect, thus energy released is related to binding energy.