Answer
$1.21m$
Work Step by Step
As $d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
$\implies d_i=(Refractive \space power-\frac{1}{d_{\circ}})^{-1}$
We plug in the known values to obtain:
$d_i=(3.6diopters-\frac{1}{0.25-0.025})^{-1}=-1.18m$
Now we can calculate the near point as
$N=1.18+0.025=1.21m$
