jmc

When we rewrite the above equation with the base numbers as sums of digit bundles we arrive at the following to work with: $$\begin{array}{rl}13_b\cdot 15_b& =243_b\Rightarrow \\ (b+3)(b+5)& =2b^2+4b+3\Rightarrow \\ b^2+8b+15& =2b^2+4b+3\Rightarrow \\ 0& =b^2-4b-12\Rightarrow \\ 0& =(b-6)(b+2).\end{array}$$ Since $b$ must be positive, the necessary base is base $\boxed{6}$.