Answer
$0.324mm$
Work Step by Step
We know that
$d_i=L-f_{eyepiece}$
$d_i=18.0-2.62=15.38cm$
Now we can find the focal length of the objective as
$f_{objective}=-\frac{d_iN}{M_{total}f_{eye\space piece}}$
We plug in the known values to obtain:
$f_{objective}=-\frac{(15.38)(25)}{(-4525)(2.62)}$
$f_{objective}=0.324mm$
