How to solve for $x$ given $x⇔A$ in a truth table?

By comparing corresponding truth values for $A$ and $(x{\iff}A)$, you can infer the corresponding truth values for $x$:

If $(x{\iff}A)=1$, then $x=A$, else $x=A'$.

Hence we can extend the truth table to include a column for $x$: $$\begin{array}{ccc|c|c} A & B & C & x{\iff}A&x\\ \hline 0 & 0 & 0 & 0&1\\ 0 & 0 & 1 & 0&1\\ 0 & 1 & 0 & 0&1\\ 0 & 1 & 1 & 1&0\\ 1 & 0 & 0 & 0&0\\ 1 & 0 & 1 & 1&1\\ 1 & 1 & 0 & 0&0\\ 1 & 1 & 1 & 0&0\\ \hline \end{array}$$ which allows us to write $$ x=A'B'C'+A'B'C+A'BC'+AB'C $$ with one term for each of the $4$ rows for which $x=1$.

As a footnote to @quasi's answer, which gives a natural and principled way of getting an answer, it is perhaps worth also noting that $x$ won't be unique.

Find one solution $x$, and any wff tautologically equivalent to $x$ will do as well. For example, $(x \land T) \leftrightarrow A$ has the same truth-table as $x \leftrightarrow A$ for any tautology $T$ at all.

Give the last column a name, say $Y$.
So, you are given $$Y = (x \iff A)$$ which is same as $$Y = (x = A)$$ so $$Y = \lnot(x \ne A)$$ Unequality of boolean values is their exclusive or: $$(p\oplus q) \equiv ((p \land\lnot q)\lor(\lnot p\land q))$$ so: $$Y = \lnot(x \oplus A)$$ hence $$\lnot Y = x \oplus A$$ and $$x = \lnot Y \oplus A$$

From the table $$Y=(A\oplus B)\land C$$ so $$x = \lnot ((A\oplus B)\land C) \oplus A$$

This expands and then contracts as $$\begin{align}x & = \lnot ((\lnot AB\lor A\lnot B)\land C) \oplus A \\ & = \lnot (\lnot ABC\lor A\lnot BC) \oplus A \\ & = ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \oplus A \\ & = \lnot((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land A \lor ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land \lnot A\\ & = (\lnot ABC\lor A\lnot BC) \land A \lor ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land \lnot A\\ & = (\lnot ABCA\lor A\lnot BCA) \lor ((A\lor\lnot B\lor\lnot C)\land \lnot A\\ & = (A\lnot BC) \lor (\lnot A\lnot B\lor\lnot A\lnot C)\\ & = A\lnot BC \lor \lnot A(\lnot B\lor\lnot C)\\ \end{align} $$

where missing operators are $\land$.