What is equivalent that logical function? 
(A+B) => (B XOR C)

=> is Implication

Options:
BC+!A+!B+!C
!B!C+ABC
B!C+A+!BC
!BC+!ABC
BC+A!B!C

! is NOT
Please tell me how it is solved. 
I do not think that this requires knowledge of discrete mathematics.
But I did not find any simple options
 A: $$(A+B) \Rightarrow (B \ XOR \ C) = \text{ (Rewrite XOR)}$$
$$(A+B) \Rightarrow (B!C + !BC) = \text{ (Rewrite} \Rightarrow )$$
$$!(A + B)+B!C + !BC = \text{ (DeMorgan)}$$
$$!A!B+B!C+!BC$$
.... which is not any of the options you are providing ... 
$BC + !A +!B+!C=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$
$!B!C+ABC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
$B!C + A +!BC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
$!BC+!ABC=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$
$BC+A!B!C=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
Are you sure you wrote this all correctly from your exercise?
EDIT
Aha! So the original is: 
$$!((A+B) \Rightarrow (B \ XOR \ C))$$
So then you get:
$$!((A+B) \Rightarrow (B \ XOR \ C)) = \text{(Rewrite XOR)}$$
$$!((A+B) \Rightarrow (B!C + !BC)) = \text{ (Rewrite} \Rightarrow )$$
$$!(!(A + B) + B!C + !BC) = \text{ (DeMorgan)}$$
$$(A + B)!(B!C + !BC) = \text{ (DeMorgan)}$$
$$(A + B)(!(B!C)!(!BC)) = \text{ (DeMorgan)}$$
$$(A + B)(!B+C)(B+!C) = \text{ (Distribution)}$$
$$A!BB+A!B!C+ACB+AC!C+B!BB+B!B!C+BCB+BC!C = \text{ (Complement)}$$
$$A0+A!B!C+ACB+A0+0B+0!C+BCB+B0 = \text{ (Annihilation)}$$
$$0+A!B!C+ACB+0+0+0+BCB+0 = \text{ (Identity)}$$
$$A!B!C+ACB+BCB = \text{ (Commutation)}$$
$$A!B!C+ABC+BBC = \text{ (Idempotence)}$$
$$A!B!C+ABC+BC = \text{ (Absorption)}$$
$$A!B!C+BC = \text{ (Commutation)}$$
$$BC+A!B!C$$
