S → A1B
A → 0A | λ
B → 0B | 1B | λWell, let's consider the leftmost derivation. We have start with S -> A1B. Since A is at the left, we use its rule. A -> λ does not work since the word would start with 1. Therefore, we apply A -> 0A two times to get 00A1B. After that A disappears to λ. We need 0 after 1, so only the rule B -> 0B works at that point. After that, we use B -> 1B and B -> λ.(a) 00101
S -> A1B -> 0A1B -> 00A1B -> 001B -> 0010B -> 00101B -> 00101.
I hope this is right. If you have concrete difficulties, post them here.