A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides $$(R_1 \neq R_2)$$, without any change in lens power then possible combination of $$R_1$$ and $$R_2$$ is :

The focal length (and therefore the power) of a thin lens is given by the Lens-Maker’s formula

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where $$\mu$$ is the refractive index of the lens material, $$R_1$$ is the radius of curvature of the first surface (positive for a convex surface facing the incident light) and $$R_2$$ is the radius of curvature of the second surface (negative for the outgoing convex surface of a bi-convex lens).

Case 1: Given bi-convex lens with equal radii
$$R_1 = +\frac16 \text{ cm}, \; R_2 = -\frac16 \text{ cm}$$ (sign taken according to the convention).

Curvature term for this lens:
$$\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{\frac16} - \frac{1}{-\frac16} = 6 - (-6) = 12 \; \text{cm}^{-1}$$

Therefore the power of the original lens is

$$P = \frac{1}{f} = (\mu - 1)\times 12 \; \text{cm}^{-1}$$ $$-(1)$$

Case 2: New convex lens with unequal radii $$R_1$$ and $$R_2$$ must satisfy the same curvature term so that its power remains unchanged:

$$\frac{1}{R_1} - \frac{1}{R_2} = 12 \; \text{cm}^{-1}$$ $$-(2)$$

Now test the four given pairs (remember the second radius must be taken negative for the second convex surface):

Option A: $$R_1 = \frac13 \text{ cm}, R_2 = -\frac13 \text{ cm}$$
$$\frac{1}{R_1} - \frac{1}{R_2} = 3 - (-3) = 6 \neq 12$$

Option B: $$R_1 = \frac15 \text{ cm}, R_2 = -\frac17 \text{ cm}$$
$$\frac{1}{R_1} - \frac{1}{R_2} = 5 - (-7) = 12$$  ✓  satisfies $$(2)$$

Option C: $$R_1 = \frac13 \text{ cm}, R_2 = -\frac17 \text{ cm}$$
$$3 - (-7) = 10 \neq 12$$

Option D: $$R_1 = \frac16 \text{ cm}, R_2 = -\frac19 \text{ cm}$$
$$6 - (-9) = 15 \neq 12$$

Only Option B preserves the value of $$\frac{1}{R_1} - \frac{1}{R_2}$$ and hence the power.

Therefore the required radii of curvature are $$\frac15 \text{ cm}$$ and $$\frac17 \text{ cm}$$, i.e. Option B.