Let the periods of the two pendula be $T_1$ and $T_2$, and their lengths be $L_1$ and $L_2$. The number of oscillations are $n_1=15$ and $n_2=10$. The total time elapsed $\Delta t$ is the same for both pendula.

$$ \Delta t = n_1 T_1 = n_2 T_2 $$

From this, we find the ratio of their periods:

$$ 15 T_1 = 10 T_2 \implies \frac{T_1}{T_2} = \frac{10}{15} = \frac{2}{3} $$

The period of a simple pendulum is given by $T = 2\pi\sqrt{L/g}$. The ratio of periods is related to the ratio of lengths:

$$ \frac{T_1}{T_2} = \frac{2\pi\sqrt{L_1/g}}{2\pi\sqrt{L_2/g}} = \sqrt{\frac{L_1}{L_2}} $$

Equating the two expressions for the ratio of periods:

$$ \sqrt{\frac{L_1}{L_2}} = \frac{2}{3} $$ $$ \frac{L_1}{L_2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} $$