Answer: $ \frac{7241379310344827586206896551}{3} = 2413793103448275862068965517.$
Solution: Represent your number as $7\cdot 10^k + a$. Then $7\cdot 10^k+a = 3(10a+7),$ and $7\cdot 10^k-29\cdot a-21 = 0.$ Therefore $29$ divides $7\cdot 10^k - 21 = 7(10^k -3)$. So $29$ divides $10^k -3$. The smallest natural $k$ such that $29$ divides $10^k -3$ is $27$ (see below how we that). From $7\cdot 10^k-29\cdot a-21 = 0$, we get that $a = 241379310344827586206896551$. Then we verify that this is indeed a valid solution.
How can we find the minimum value of $k$ s.t. $\mathbf{10^k \equiv 3 \pmod{29}}$ ? This problem is known as the Discrete Logarithm Problem. It is believed that in general there is no efficient algorithm for it. Since all parameters are very small in this problem, the easiest solution is just to consecutively try $k$ from $1$ to $29$ until we find $k$. In order to compute $10^{k+1} \mathrm{\ mod \ } 29$, we just take $10^{k} \ \mathrm{mod}\ 29$ multiply it by $10$ and divide by $29$. We get, \begin{align*} &{}\ 10^{1} \equiv10, && 10^{2} \equiv 10 \cdot {10} \equiv13, && 10^{3} \equiv 10 \cdot {13} \equiv14, && 10^{4} \equiv 10 \cdot {14} \equiv24,\\ &{}\ 10^{5} \equiv 10 \cdot {24} \equiv8, && 10^{6} \equiv 10 \cdot {8} \equiv22, && 10^{7} \equiv 10 \cdot {22} \equiv17, && 10^{8} \equiv 10 \cdot {17} \equiv25,\\ &{}\ 10^{9} \equiv 10 \cdot {25} \equiv18, && 10^{10} \equiv 10 \cdot {18} \equiv6, && 10^{11} \equiv 10 \cdot {6} \equiv2, && 10^{12} \equiv 10 \cdot {2} \equiv20,\\ &{}\ 10^{13} \equiv 10 \cdot {20} \equiv26, && 10^{14} \equiv 10 \cdot {26} \equiv28, && 10^{15} \equiv 10 \cdot {28} \equiv19, && 10^{16} \equiv 10 \cdot {19} \equiv16,\\ &{}\ 10^{17} \equiv 10 \cdot {16} \equiv15, && 10^{18} \equiv 10 \cdot {15} \equiv5, && 10^{19} \equiv 10 \cdot {5} \equiv21, && 10^{20} \equiv 10 \cdot {21} \equiv7,\\ &{}\ 10^{21} \equiv 10 \cdot {7} \equiv12, && 10^{22} \equiv 10 \cdot {12} \equiv4, && 10^{23} \equiv 10 \cdot {4} \equiv11, && 10^{24} \equiv 10 \cdot {11} \equiv23,\\ &{}\ 10^{25} \equiv 10 \cdot {23} \equiv27, && 10^{26} \equiv 10 \cdot {27} \equiv9, && \boxed{\mathbf{10^{27} \equiv 10 \cdot {9} \equiv3}} \end{align*} (all computations are modulo $29$).