The magnitudes of power of a biconvex lens (refractive index 1.5)...

Let $R_1$ and $R_2$ be the radii of the front and back surfaces of the biconvex lens. The power of the biconvex lens is $P_1 = (1.5 - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$.
The plano-concave lens has a flat surface and a concave surface of radius $R_2$ (matching the back surface of the biconvex lens). Its power is $P_2 = (1.7 - 1)\left(\frac{1}{R_2}\right) = \frac{0.7}{R_2}$.
Since the magnitudes of power are equal: $0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$
Simplifying: $\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2}$, which gives $\frac{0.5}{R_1} = \frac{0.2}{R_2}$
Therefore: $\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$