= Solution
{c}
If the <prime number> $p$ divides $2^q-1$, then $p\ne2$, because $2^q-1$ is odd. Thus the residue of $2$ is an element of the <multiplicative group of a finite field> $\mathbb Z_p$, and $2^q\equiv1\pmod p$.
Let $d$ be its <order of a group element>. The <integer division> writes $q=ad+r$ with $0\leq r<d$. Then $1\equiv2^q\equiv(2^d)^a2^r\equiv2^r\pmod p$, so minimality of $d$ gives $r=0$. Hence $d$ divides $q$. Since $q$ is a <prime number>, $d=1$ or $d=q$. The former would give $2\equiv1\pmod p$, which would mean $p$ divides $1$, an impossibility. Consequently
$$
\boxed{\operatorname{ord}_p(2)=q.}
$$
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