Step 1: Identify the eye defect.
The person cannot see things closer than $100$ cm, but a normal near point is $25$ cm. So their near point is pushed far away. This is hypermetropia, and it needs a converging lens.
Step 2: Decide what the lens must do.
The lens should take an object held at the normal reading distance $25$ cm and form a virtual image at the person's own near point $100$ cm. Using sign convention: \[ u = -25\ \text{cm}, \qquad v = -100\ \text{cm} \]
Step 3: Write the lens formula.
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-100} - \frac{1}{-25} \]
Step 4: Simplify.
\[ \frac{1}{f} = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100} \] So $f = \dfrac{100}{3}$ cm.
Step 5: Convert focal length to power.
Power in dioptres uses focal length in metres, or simply $P = \dfrac{100}{f(\text{cm})}$: \[ P = \frac{100}{100/3} = +3\ \text{D} \]
Step 6: State the answer.
A converging lens of power: \[ \boxed{P = +3\ \text{D}} \]