If the coefficients $a_i,b_i,c_i$ are not constrained to be non-negative, $a_2$ can be made as large as possible.
Consider $b_1 = b_2 = b_3 = \frac{500}{3}$. Let $a_2 =M$ be any positive number. $a_1 = \frac{5}{3}M$. Also, let $c_3 = -1$. Then, $c_1 = c_2 = \frac{18+5M}{8M}$ will satisfy all the equations. If $M$ is sufficiently large$(>\frac{3}{8}), a_3<0$. So, the product of all terms will be positive.
So, $a_2$ can be made as large as wanted. In other words, it is unbounded.
If you assume that all the $a_i,b_i,c_i$ are non-negative, the problem is straight-forward.
Define new variables $x_1 = a_1b_1c_1,x_2 = a_2b_2c_2,x_3 = a_3b_3c_3$ in the first relation. Then, solve the simple LP that arises.
Try any online solver or matlab or mathematica for a sloution. You need not know how to solve a LP to get a solution.