If $T_n$ is a Cauchy sequence in $\mathcal L (X,B)$, then for any chosen $x \in X$, $T_n(x)$ is a Cauchy sequence in $B$. Since $B$ is complete, this sequence converges.
Define a new linear operator $T : X \to B$ mapping $x \mapsto \lim_{n \to \infty} T_n(x)$.
Now prove that $T_n \to T$ in the operator norm. Let $\epsilon > 0$. The Cauchy property tells you that there exists an $N$ such that
$$ m,n > N \implies \sup_{||x || \leq 1} ||T_n(x) - T_m (x) || < \epsilon.$$
Take the limit $m \to \infty$.
[If your notation $\mathcal L (X, B)$ refers to bounded operators, then you also need to prove that $T$ is bounded. I'll leave you to do that if required.]
Here is a way to show this when you consider the field of real numbers and this for all norms.
Consider $E$ a finite dimensional vector space over $\mathbb{R}$ and a basis $(e_1, ..., e_n)$ of $E$. It means that for all $x\in E$ you have $x = \sum_{i=1}^{n} x_{i}e_i $ where the $x_i$ are real numbers.
Now we need to endow E with a norm, first recall that all norms are equivalents in a finite dimensional normed space it will be usefull at some points.
The norm we will consider is $\lVert x\rVert_{\infty} = max_{1\leq i\leq n}\lvert x_i\rvert $. Thus $(E,\lVert\rVert_{\infty})$ is a finite dimensional normed space.
Now consider a Cauchy sequence in $E$, namely $(x^{p})_{p\in\mathbb{N}}$, it means that
$\forall\epsilon>0,\exists N\in\mathbb{N} : n,m\geq N\implies\lvert x_i^{n} -x_i^{m}\rvert\leq\lVert x^{n} - x^{m}\rVert_{\infty} = max_{1\leq i \leq n}\lvert x_{i}^{n} - x_{i}^{m}\rvert\leq\epsilon$
However for $i\in\{1,...,n\}$ we have that $(x_i^{p})$ is a sequence of real numbers and we have shown that it is Cauchy !
Thus for all $i\in\{1,..n\}$ there exists $x_i\in\mathbb{R}$ such that $\forall\epsilon>0, \exists N_{i}\in\mathbb{N} : p\geq N_i\implies \lvert x_i^{p} -x_i\rvert\leq\epsilon$ (since $\mathbb{R}$ is complete)
Now, a candidate we should consider as a limit of $(x^p)_{p\in\mathbb{N}}$ is the vector whose coordinate corresponds to the $x_i$'s that is $x = \sum_{i=1}^{n}x_ie_i$.
Fix $\epsilon>0$ and $N = max_{1\leq i\leq n}(N_i)$, then we have that
$p\geq N\implies max_{1\leq i\leq n}\lvert x_i^p - x_i\rvert = \lVert x^p - x\rVert_{\infty}\leq\epsilon$ which shows the convergence of $(x^p)_{p\in\mathbb{N}}$ to $x\in E$.
Since $E$ is a finite dimensional normed space, all norms on this space are equivalents which means that the convergence is not altered by a change of norm (this is already explained in the comments I think)
I hope this is clear !
Best Answer
To provide an answer without using the norm completion, consider a norm-Cauchy sequence $x_n$ that is weakly convergent to some $x\in X$. By considering $x_n-x$ we may assume that $x_n$ converges weakly to $0$.
Now suppose that $x_n$ is not norm-convergent to $0$, i.e. there is some $\epsilon>0$ with $\|x_n\|>\epsilon$ for infinitely many $n$. By rescaling the sequence with $1/\epsilon$ and by throwing away some terms we may assume $\|x_n\|>1$ for all $n$. Additionally we may pass to a subsequence to get: $$\|x_n-x_2\|≤\frac12.$$ Now let $f$ be some dual element so that $f(x_2)=\|x_2\|≥1$ and $\|f\|=1$. Then for any $n≥2$ you have $f(x_n) = f(x_2)+f(x_2-x_n) ≥\|x_2\|-\|x_2-x_n\| ≥ 1-2^{-1}=1/2$. As consequence $f(x_n)\not\to0$, contradicting that $x_n\to0$ weakly.