Build a regular grammar for a regular language
The expression always starts with 1a. There are two options, the third character in the string is a or not. If it is a, then you don't have the obligation of following 'it contains at least one a'. Otherwise, you need to. Thus the regular expression is:

    1aa(1+a+b)*+1a(1+a+b)*aa(1+a+b)* .


The conversion from regular expression to automata to regular grammar is standard. One possible algorithm is given here: https://math.stackexchange.com/questions/574571/build-regular-grammar-from-regular-expression

Here is an intuitive way to convert a regular expression to a right linear grammar:


Convert the regular expression to an NFA. This is standard (see page 102 of Automata Theory, language and computation, 3rd ed).
To convert the NFA to a right linear grammar, take the states to be non-terminal and the alphabets to be terminal. For a transition of the form S -&gt; S' via a, add the production rule S -&gt; aS'. Please note that you need to take care of the final states and epsilon transitions separately.


You can also convert the NFA to a DFA (again, standard subset construction method, see the aforementioned book) and then the method will be simpler.