Completion of borel sigma algebra with respect to Lebesgue measure

From the definition of the outer measure $\lambda^{*}$, you can show that if $A\in \mathcal{L}'$ then there's a $G_{\delta}$ set $B$ so that $A\subseteq B$ and $\lambda^{*}(B\setminus A)=0$. After that, the answer to this question is an easy yes.

Just filling the details of Michael's answer:

If $A\in\mathcal{L}'$, then consider the family $\mathcal{C}$ of every sequence of products of open intervals $\{R_i\}_{i=1}^{\infty}$ such that $A\subset\bigcup_{i=1}^{\infty}R_i$. Now, by definition of the Lebesgue outer measure, $\lambda^*(A):=\inf\left\{\sum_{i=1}^{\infty}\mathrm{vol}(R_i):\{R_i\}_{i=1}^{\infty}\in\mathcal{C}\right\}$. So it is easy to show that there is a decreasing sequence of sets $\{U_j\}_{j=1}^{\infty}$ being $U_j=\bigcup_{i=1}^{\infty}R_i$ for some $\{R_i\}_{i=1}^{\infty}\in\mathcal{C}$ such that $A\subset B:=\lim_{j\to\infty}U_j:=\bigcap_{j=1}^{\infty}U_j\in\mathcal{B}$ holds that $\lambda^*(B)=\lambda^*(A)$.

Using that $A\in\mathcal{L}'$ it follows that $\lambda^*(B)=\lambda^*(A\cap B)+\lambda^*(B\setminus A)$, so $\lambda^*(B\setminus A)=0$.

Knowing this, we have to proof that given any $A\in \mathcal{L}'$, then $A=A_1\cup A_2$, being $A_1\in \mathcal{B}$ and $A_2\subset N\in\mathcal{B}$ such that $\lambda(N)=0$. It is easy to show that $\mathbb{R}^n\setminus A=(B\setminus A)\cup(\mathbb{R}^n\setminus B)$, which fulfill all our requirements. So, as every $A\in\mathcal{L'}$ is the complementary of another set in $\mathcal{L'}$, we have won, and $\mathcal{L}=\mathcal{L'}$.