Presumably, $\gcd(a,29)=1$; else, $a\equiv 0\pmod{29}$, and all your incongruences hold, and there is no order as Sivaram points out.
From Fermat's Little Theorem, you know that $a^{28}\equiv 1\pmod{29}$.
Now, what is the smallest positive integer $j$ such that $a^j\equiv 1\pmod{29}$? It is at most $28$. If we divide $28$ by $j$ with remainder, $28 = qj + r$, $0\leq r\lt j$, we have that
$$a^r = a^{28-qj} \equiv a^{28}(a^{j})^{-q} \equiv 1 \pmod{29}.$$
Since $j$ is the smallest positive integer such that $a^j\equiv 1 \pmod{29}$, but $r$ is nonnegative and smaller, you must have $r=0$. (This is a standard argument, shows up all the time; it is really the fact that the set of exponents $r$ for which $a^r\equiv 1 \pmod{29}$ form an ideal of the integers). So $j$ divides $28$. That is, the order must divide $28$.
So what can $j$ be? It must be positive, it must divide $28=2^2\times 7$. So it must be either $1$, $2$, $4$, $7$, $14$, or $28$. Given the information you have, what must $j$ be?