 Solution

For each fixed input length, take the complete truth table of the language. For every accepted string $z\in \{0,1{\}}^n$, form the mintermAn OR of these minterms equals the indicator function, since $M_z(x)=1$ precisely at $x=z$. This is the truth-table upper bound for circuit size.

Generate the $n$ negated input wires once and share them. With $m\le 2^n$ accepted strings, use at most $m(n-1)$ binary AND gates and $m-1$ binary OR gates, besides the negations. Empty truth tables use a constant-zero Boolean circuit; length zero is handled by a constant Boolean circuit. ThusThis is an existence bound for a nonuniform circuit family, even when the language is undecidable. It supplies no algorithm for computing the truth tables of an arbitrary language.