This is standard, and the answer has been been indicated in the comments. Recall that in a normed space $X,$ the triangle inequality easily yields that $| \|x\| - \| y \| |\leq \|x-y\|$
for all $x,y \in X.$

Now let's turn to your dense subspace $V$ of the Banach space $B.$ Take an element $b \in B.$ There is a sequence $(v_{n})$ of elements of $V$ such that $v_{n} \to b$ (with respect to the norm on $B).$ Then by the above remark, the sequence $( \|v_{n} \|)$ is a Cauchy sequence of real numbers whose limit is the real number $\| b \|$. Hence $\| b \|$ is uniquely specified in terms of $ \| \|_{V} $ since you assume that $\| \|$ and $\| \|_{V}$ agree on $V.$ Also, $\|b\|$
is the same as would be assigned if considering $b$ as an element of the completion of $V.$
Hence $B$ embeds isometrically as a susbpace of the completion of $V.$ However, it is clearly dense in that completion, as $V$ already is, and it is a closed subspace of that completion, since any Cauchy sequence of elements of $V$ has a limit which lies in $B.$ Hence $B$ is indeed isometrically isomorphic to the completion of $V.$

acompletion of $V$. Note thatthecompletion does not make sense. – Jochen Wengenroth Jan 11 '13 at 9:44acompletion. All the other completions can be obtained by an isometry. What do you mean by "the norms are identical"? – Davide Giraudo Jan 11 '13 at 9:44particular constructionof a completion, e.g. via equivalence classes of Cauchy seuences? – Yemon Choi Jan 11 '13 at 10:07