A nearsighted person is unable to see objects beyond 25 cm ....

For viewing distant objects, the object distance is $u = -\infty$.

The far point of the nearsighted eye is at $25 \text{ cm}$ from the eye.

Since the lens is placed $1 \text{ cm}$ from the eye, the image must be formed at the far point. Thus, the image distance from the lens is $v = -(25 - 1) \text{ cm} = -24 \text{ cm} = -0.24 \text{ m}$.

Using the lens formula:

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\dfrac{1}{f} = \dfrac{1}{-0.24} - \dfrac{1}{-\infty}$

$f = -0.24 \text{ m}$

The power of the lens is given by:

$P = \dfrac{1}{f} = \dfrac{1}{-0.24} = -4.166 \text{ D} \approx -4.2 \text{ D}$