1GB = $2^{30}B$
DB = Disk Block
DBA length or size = 32 bits
= 4 B
#DBAs that can be stored per DB = DB size/ DBA size
= $\frac{2^{12}}{2^2}$
= $2^{10}$
TOTAL file size
= $(\#DirectPtr + \sum_i(\#Ptr_i * \#DBAs)^{i} ) * DB size$
Where i = indirectionLevel (single = 1, double = 2, etc)
Each direct pointer points to a single DB and for indirect pointer it depends on the indirectionLevel (single, double, triple or etc) to point to that many DBs. Thus we get max file size by multiplying total DB pointers with DBsize.
= $(12 + \sum (1*2^{10})^1 \ + (1*2^{10})^2 )*2^{12}\ B$
=$4. 004\ GB$ or
= 4.0(ANS)