As the series lies between 100 & 1000 , the desired series of palindrome will have only 3 digit nos.
Start from the 1 series, keeping the 1st & last digit as 1 (1st no. will be 101) ,
we can have 10 (0,1,2...9) different ways for choosing the 2nd digit .
Repeating the same process for 2 series (where 1st & last digit is 2) and so on...we will always have 10 different nos. for each series.
Therefore , adding the nos. till the 9 series (last no. being 999) = 10*9 = 90

Stage 1 - Selecting from 9 digits I get (0-9)
Stage 2 - Selecting from 10 digits - Little lost here . Since one digit is already selected in stage 1 - Shouldnt it be 9 digits to select from ?

There are two situations:

1. xyx (ex: 101, 121..)
2. xxx (ex: 111, 2222..)

For the first case:

xyx: The hundredth position can be taken by any one among the nine digits..(1,2,3...9)
The tenth position can be any one among the nine numbers (0,1,2..9 except the digit which we have taken in the hundredth position)
The unit's place should be the same digit which we assigned in hundredth position.
Hence this can be arranged in 9 * 9 * 1 = 81

For the second case:

xxx: The hundredth position can be taken by any one among the nine digits..(1,2,3...9)
The tenth and the units's place should take the same digit which we assigned in hundredth position.
Hence this can be arranged in 9 * 1 * 1 = 9

The total number of palindromes between 100 and 1000 = 81 + 9 =90.

ANSWER: 90