Given,

Refractive index of the lens material, ^aμ_g =1.5

Focal length of the lens in air, f_air = 15 cm

Using lens makers formula-

1/f_air = (^aμ_g-1) [1/R₁ -1/R₂]

1/15 = (1.5-1)[1/R₁ -1/R₂]

[1/R₁ -1/R₂] = 1/7.5

Refractive index of liquid, ^aμ_L =1.7

Refractive index of lens glass with respect to liquid, ^Lμ_g=^aμ_g /^aμ_L

 ^Lμ_g= 1.5/1.7 = 0.88

Let the focal length of the lens in liquid is-

1/f_liquid = (^Lμ_g-1)[1/R₁ -1/R₂]

1/f_liquid = (0.88-1)[1/7.5]

= -0.015

So,f_liquid = -66.67 cm

From the lens makers formula

1/f = (n-1)[1/R₁ -1/R₂]

Where n, is the refractive index of material of the lens with respect to refractive to refractive index of the medium.

n = n_lens /n _medium

If we take,

 refractive index of medium = refractive index of lens material

Then, n= 1

And 1/f =0

So power of lens is zero.

In this condition, the lens disappears in liquid.

So, to make the lens invisible in liquid, the refractive index of lens material should be equal to the refractive index of liquid.

Which is 1.7 in this question.