Answer: (B).
Lets go over each of the options with case method where $b = True$.
A) $a \vee b → b \wedge c$
$=> (a \vee True) → (True \wedge c)$
$=> (True) → (c)$
$=> c$
Option A will not be Tautology if $c=False$.
B) $a \wedge b → b \vee c$
$=> (a \wedge True) → (True \vee c)$
$=> (a) → True$
$=> True$ (Anything implies True is always True)
Option B is Tautology.
C) $a \vee b → (b → c)$
$=> a \vee True → (True → c)$
$=> (True) → (c) $
$=> c $
Option C will not be Tautology if $c=False$.
C) $a → b → (b → c)$
$=> a → (True → (True → c))$
$=> a → (True → (c))$
$=> a → c$
Option D will not be Tautology if $a=True$ and $c=False$