= Solution
Under <duality of sequence spaces>, $a=(a_n)\in\ell_1$ represents $f_a(x)=\sum a_nx_n$ on $c_0$, with <operator norm> $\|f_a\|=\sum|a_n|$. Here $c_0$ is the <space of sequences converging to zero>, equipped with its supremum <norm>.
If $\|x\|_\infty\leq1$ and $f_a(x)=\sum|a_n|$, then
$$
0=\sum_{n\geq1}(|a_n|-a_nx_n).
$$
Every summand is nonnegative, so $x_n=\operatorname{sgn}(a_n)$ whenever $a_n\ne0$. Infinitely many such coordinates would give infinitely many $|x_n|=1$, contradicting $x_n\to0$. Therefore $a$ has finite support.
Conversely, if $a$ has finite support, choose $x_n=\operatorname{sgn}(a_n)$ on that support and $x_n=0$ elsewhere. Then $x\in c_0$, $\|x\|_\infty\leq1$, and $f_a(x)=\|f_a\|$. The zero functional is included by taking $x=0$. Thus
$$
\boxed{f_a\text{ attains its norm on }B_{c_0}\iff a\in c_{00}.}
$$
This is <norm attainment on the vanishing sequence space>.
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