A formula in Reed Muller normal form consists of conjunctions of variables. These conjunctions are connected by xor. Hence, the first formula does not meet that syntactic requirement, the second one does.

My solution looks like this:

(c ∧ ¬b) ∨ (¬c ∧ a)

// note that there is no assignment that satisfies both (c ∧ ¬b) and (¬c ∧ a) at the same time, so ∨ and ⊕ make no difference here.

= (c ∧ ¬b) ⊕ (¬c ∧ a)

// now we express negation by ⊕ 1

= (c ∧ (1 ⊕ b)) ⊕ ((1 ⊕ c) ∧ a)

// now we multiply out

= c ⊕ b ∧ c ⊕ a ⊕ c ∧ a